test-equations.txt

\(\pi\)
----
\(\frac{22}{7}\)
----
\(ℕ\)
----
\(\{1; \; 2; \; 3; \; \ldots\}\)
----
\(ℕ_0\)
----
\(\{0; \; 1; \; 2; \; 3; \; \ldots\}\)
----
\(ℤ\)
----
\(\{\ldots; \; -3; \; -2; \; -1; \; 0; \; 1; \; 2; \; 3; \; \ldots\}\)
----
\(ℚ\)
----
\(\frac{a}{b}\)
----
\(a\)
----
\(b\)
----
\(b\ne 0\)
----
\(ℚ'\)
----
\(ℝ\)
----
\(ℝ'\)
----
\(\sqrt{-25}; \; \sqrt[4]{-1}; \; -\sqrt{-\frac{1}{16}}\)
----
\(ℕ\)
----
\(ℕ_0\)
----
\(ℤ\)
----
\(ℚ\)
----
\(ℚ'\)
----
\(ℝ\)
----
\(ℝ'\)
----
\(\sqrt{7}\)
----
\(ℝ; ℚ'\)
----
\(ℝ; ℚ\)
----
\(16\frac{2}{5}\)
----
\(ℝ; ℚ\)
----
\(\sqrt{6\frac{1}{4}}\)
----
\(ℝ; ℚ\)
----
\(ℝ; ℚ; ℤ;  ℕ_0\)
----
\(2\pi\)
----
\(ℝ' ℚ'\)
----
\(ℝ; ℚ\)
----
\(\frac{1-\sqrt{2}}{2}\)
----
\(ℝ; ℚ'\)
----
\(-\sqrt{-3}\)
----
\(ℝ'\)
----
\((\pi)^2\)
----
\(ℝ; ℚ'\)
----
\(-\frac{9}{11}\)
----
\(ℝ; ℚ\)
----
\(\sqrt[3]{-8}\)
----
\(ℝ; ℚ; ℤ\)
----
\(\frac{22}{7}\)
----
\(ℝ; ℚ\)
----
\(ℝ; ℚ'\)
----
\(ℝ; ℚ\)
----
\(\sqrt[5]{-32}\)
----
\(ℝ; ℚ; ℤ\)
----
\[a^n = a \times a \times a \times \ldots \times a \quad (n \text{ times}) \qquad \left(a \in ℝ, n \in ℕ\right)\]
----
\(2 \times 2 \times 2 \times 2 = 2^4\)
----
\(k^6 = k \times k \times k \times k \times k \times k\)
----
\(x^2\)
----
\(x\)
----
\(y^3\)
----
\(y\)
----
\(k^6\)
----
\(k\)
----
\({a}^{0}=1\phantom{\rule{28.45274pt}{0ex}}\)
----
\(a \ne 0\)
----
\(0^0\)
----
\({a}^{-n}=\frac{1}{{a}^{n}}\phantom{\rule{14.22636pt}{0ex}}\)
----
\(a \ne 0\)
----
\(\frac{1}{0}\)
----
\(5^{-2} = \dfrac{1}{5^2} = \dfrac{1}{25}\)
----
\((-36)^0 x = (1)x = x\)
----
\(\dfrac{7p^{-1}}{q^{3}t^{-2}} = \dfrac{7t^2}{pq^3}\)
----
\({a}^{m} \times {a}^{n}={a}^{m+n}\)
----
\(\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}\)
----
\({\left(ab\right)}^{n}={a}^{n}{b}^{n}\)
----
\({\left(\frac{a}{b}\right)}^{n}=\frac{{a}^{n}}{{b}^{n}}\)
----
\({\left({a}^{m}\right)}^{n}={a}^{mn}\)
----
\(a > 0\)
----
\(b > 0\)
----
\(m, n \in ℤ\)
----
\(5(m^{2t})^p \times 2(m^{3p})^t\)
----
\(\dfrac{8k^3x^2}{(xk)^2}\)
----
\(\dfrac{2^2 \times 3 \times 7^4}{(7 \times 2)^4}\)
----
\(3(3^b)^a\)
----
\(5(m^{2t})^p \times 2(m^{3p})^t = 10m^{2pt + 3pt} = 10m^{5pt}\)
----
\(\dfrac{8k^3x^2}{(xk)^2} =  \dfrac{8k^3x^2}{x^2k^2} = 8k^{(3-2)}x^{(2-2)} = 8k^1x^0 = 8k\)
----
\(\dfrac{2^2 \times 3 \times 7^4}{(7 \times 2)^4} = \dfrac{2^2 \times 3 \times 7^4}{7^4 \times 2^4} = 2^{(2-4)} \times 3 \times 7^{(4-4)} = 2^{-2} \times 3 = \frac{3}{4}\)
----
\(3(3^b)^a = 3 \times 3^{ab} = 3^{ab + 1}\)
----
\(\dfrac{3^m - 3^{m+1}}{4 \times 3^m - 3^m}\)
----
\[\dfrac{3^m - 3^{m+1}}{4 \times 3^m - 3^m} = \dfrac{3^m - (3^{m} \times 3)}{4 \times 3^m - 3^m}\]
----
\[\begin{aligned}
										&= \dfrac{3^m(1 - 3)}{3^m(4 - 1)} 
										\end{aligned}\]
----
\[\begin{aligned}
										&= \frac{1 - 3}{4 - 1}  \\
										&= - \frac{2}{3}
										\end{aligned}\]
----
\(4 \times 4^{2a} \times 4^2 \times 4^a\)
----
\(\dfrac{3^2}{2^{-3}}\)
----
\((3p^5)^2\)
----
\(\dfrac{k^2k^{3x-4}}{k^x}\)
----
\((5^{z-1})^2+5^z\)
----
\((\frac{1}{4})^0\)
----
\((x^2)^5\)
----
\(\left( \frac{a}{b} \right)^{-2}\)
----
\((m+n)^{-1}\)
----
\(2(p^t)^s\)
----
\(\dfrac{1}{\left(\frac{1}{a}\right)^{-1}}\)
----
\(\frac{k^{0}}{k^{-1}}\)
----
\(\dfrac{-2}{-2^{-a}}\)
----
\(\dfrac{-h}{(-h)^{-3}}\)
----
\(\left( \dfrac{a^2b^3}{c^3d} \right)^2\)
----
\(10^{7}(7^{0}) \times 10^{-6}(-6)^{0}-6\)
----
\(m^3n^2 \div nm^2 \times \frac{mn}{2}\)
----
\((2^{-2}-5^{-1})^{-2}\)
----
\((y^2)^{-3} \div \left( \frac{x^2}{y^3} \right)^{-1}\)
----
\(\dfrac{2^{c-5}}{2^{c-8}}\)
----
\(\dfrac{2^{9a} \times 4^{6a} \times 2^2}{8^{5a}}\)
----
\(\dfrac{20t^5p^{10}}{10t^4p^9}\)
----
\(\left( \dfrac{9q^{-2s}}{q^{-3s}y^{-4a-1}} \right)^2\)
----
\(r^n = a\)
----
\(r = \sqrt[n]{a} \quad (n \geq 2)\)
----
\(a^{\frac{1}{n}} = \sqrt[n]{a}\)
----
\(a^{-\frac{1}{n}} = (a^{-1})^{\frac{1}{n}} = \sqrt[n]{\frac{1}{a}}\)
----
\(a^{\frac{m}{n}} = (a^{m})^{\frac{1}{n}} = \sqrt[n]{a^m}\)
----
\(a > 0\)
----
\(r > 0\)
----
\(m,n \in ℤ\)
----
\(n \ne 0\)
----
\(\sqrt{25} = 5\)
----
\(\sqrt[3]{8} = 2\)
----
\(\sqrt[5]{32} = 2\)
----
\(n\)
----
\(\sqrt[4]{16} = 2\)
----
\(2 \times 2 \times 2 \times 2 = 16\)
----
\(\sqrt[4]{-16}\)
----
\((-2) \times (-2)  \times (-2) \times (-2) \ne -16\)
----
\(n\)
----
\(\sqrt[3]{27} = 3\)
----
\(3 \times 3 \times 3 = 27\)
----
\(\sqrt[3]{-27} = -3\)
----
\((-3) \times (-3) \times (-3) = -27\)
----
\(\left(-2\right)^2 = 4\)
----
\(2^2 = 4\)
----
\(\sqrt{12}\)
----
\(\sqrt[5]{25}\)
----
\(18^\frac{1}{2}\)
----
\(4^\frac{3}{2}\)
----
\((-81)^\frac{1}{2}\)
----
\(18^\frac{1}{2} = \sqrt{18}\)
----
\(4^\frac{3}{2} = (4^3)^{\frac{1}{2}} = \sqrt{4^3} = \sqrt{64} = 8\)
----
\((-81)^\frac{1}{2} = \sqrt{-81} =\)
----
\[\left(\frac{5}{4^{-1}-9^{-1}}\right)^{\frac{1}{2}}\]
----
\[\left( \dfrac{5}{\frac{1}{4} - \frac{1}{9}} \right)^{\frac{1}{2}}\]
----
\[\begin{align}
		  								&= \left( \dfrac{5}{\frac{9-4}{36}} \right)^{\frac{1}{2}} \\
		  								&= \left( \dfrac{5}{\frac{5}{36}} \right)^{\frac{1}{2}} \\
		  								&= \left( 5 \div \frac{5}{36} \right)^{\frac{1}{2}} \\
		  								&= \left( 5 \times \frac{36}{5} \right)^{\frac{1}{2}} \\
		  								&= (36)^{\frac{1}{2}} 
		   								\end{align}\]
----
\[\begin{align}
		  								&= \sqrt{36}\\
		  								&= 6
		   								\end{align}\]
----
\(\sqrt{49}\)
----
\(\sqrt{36^{-1}}\)
----
\(\sqrt[3]{6^{-2}}\)
----
\(\sqrt[3]{-\dfrac{64}{27}}\)
----
\(\sqrt[4]{(16x^4)^3}\)
----
\({s}^{\frac{1}{2}}÷{s}^{\frac{1}{3}}\)
----
\({\left(64{m}^{6}\right)}^{\frac{2}{3}}\)
----
\(\dfrac{12m^{ \frac{7}{9}}}{8m^{- \frac{11}{9}}}\)
----
\(x\)
----
\(x\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}\)
----
\(\sqrt[n]{a}\sqrt[n]{b} = \sqrt[n]{ab}\)
----
\(\sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}\)
----
\(\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}\)
----
\(\sqrt[n]{a^m} = a^{\frac{m}{n}}\)
----
\(\left( \sqrt[n]{a} \right)^m = a^{\frac{m}{n}}\)
----
\(\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{ab}\)
----
\(\sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}\)
----
\[\begin{align}
		    \sqrt[n]{a} \times \sqrt[n]{b}
		    &= a^{\frac{1}{n}} \times b^{\frac{1}{n}} \\
		    &= (ab)^{\frac{1}{n}} \\
		    &= \sqrt[n]{ab}
		    \end{align}\]
----
\[\begin{align}
		    \sqrt[n]{\frac{a}{b}}
		    &= \left( \frac{a}{b} \right)^{\frac{1}{n}} \\
		    &= \dfrac{a^{\frac{1}{n}}}{b^{\frac{1}{n}}} \\
		    &= \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}
		    \end{align}\]
----
\(\sqrt{2} \times \sqrt{32} = \sqrt{2 \times 32} = \sqrt{64} = 8\)
----
\(\dfrac{\sqrt[3]{24}}{\sqrt[3]{3}} = \sqrt[3]{\dfrac{24}{3}} = \sqrt[3]{8} = 2\)
----
\(\sqrt{\sqrt{81}} = \sqrt[4]{81} = \sqrt[4]{3^4} = 3\)
----
\(\sqrt[m]{a}\)
----
\(\sqrt[n]{b}\)
----
\(m=n\)
----
\(\sqrt{\frac{1}{3}}\)
----
\(-\sqrt{61}\)
----
\(m = n = 2\)
----
\(\sqrt[3]{5} \text{ and } \sqrt[5]{7y^3}\)
----
\(m \ne n\)
----
\(\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}\)
----
\(\sqrt{50}\)
----
\[\begin{align}
		  								\sqrt{50} &= \sqrt{5 \times 5 \times 2} \\
		  								          &= \sqrt{5^2 \times 2}
		   								\end{align}\]
----
\(\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}\)
----
\[\begin{align}
		  								&= \sqrt{5^2} \times \sqrt{2} \\  
		  								&= 5 \times \sqrt{2} \\ 			
		  								&= 5\sqrt{2}					
		   								\end{align}\]
----
\(\sqrt{6}, \sqrt[3]{30} \text{ and } \sqrt[4]{42}\)
----
\(\sqrt[3]{54}\)
----
\[\begin{align}
		  								\sqrt[3]{54} &= \sqrt[3]{3 \times 3 \times 3 \times 2} \\
		  								             &= \sqrt[3]{3^3 \times 2}
		   								\end{align}\]
----
\(\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}\)
----
\[\begin{align}
		  								&= \sqrt[3]{3^3} \times \sqrt[3]{2} \\  
		  								&= 3 \times \sqrt[3]{2} \\ 			
		  								&= 3\sqrt[3]{2}					
		   								\end{align}\]
----
\(\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}\)
----
\[\begin{align}
		    &= \left( \sqrt{7^2} \times \sqrt{3} \right) + \left( \sqrt{6^2} \times \sqrt{3} \right)\\  
		    &= \left( 7 \times \sqrt{3} \right) +  \left( 6 \times \sqrt{3} \right) \\   			
		    &= 7\sqrt{3} + 6\sqrt{3}			
		    \end{align}\]
----
\[13 \sqrt{3}\]
----
\(\left( \sqrt{20} - \sqrt{5} \right)^2\)
----
\[\left( \sqrt{20} - \sqrt{5} \right)^2 = \left( \sqrt{4 \times 5} - \sqrt{5} \right)^2\]
----
\(\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}\)
----
\[\begin{align}
		    &= \left( \sqrt{4} \times \sqrt{5} - \sqrt{5} \right)^2	 \\  
		    &= \left( 2 \times \sqrt{5} - \sqrt{5} \right)^2	 \\  			
		    &= \left( 2\sqrt{5} - \sqrt{5} \right)^2
		    \end{align}\]
----
\[\begin{align}  			
		    &= \left( \sqrt{5} \right)^2	 \\
		    &= 5	 
		    \end{align}\]
----
\(\sqrt{75} \times \sqrt[3]{(48)^{-1}}\)
----
\[\begin{align}
		    \sqrt{75} \times \sqrt[3]{(48)^{-1}} 	&= \sqrt{25 \times 3} \times \sqrt[3]{\frac{1}{48}} \\		
		    &= \sqrt{25 \times 3} \times \frac{1}{\sqrt[3]{8 \times 6}} 
		    \end{align}\]
----
\(\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}\)
----
\[\begin{align}
		    &= \sqrt{25} \times \sqrt{3} \times \frac{1}{\sqrt[3]{8} \times \sqrt[3]{6}} \\  
		    &= 5 \times \sqrt{3} \times \frac{1}{2 \times \sqrt[3]{6}}  					
		    \end{align}\]
----
\[\begin{align}  			
		    &= 5\sqrt{3} \times \frac{1}{2\sqrt[3]{6}} \\
		    &= \frac{5\sqrt{3}}{2\sqrt[3]{6}} 
		    \end{align}\]
----
\(\sqrt[3]{16} \times \sqrt[3]{4}\)
----
\(\sqrt{a^2b^3} \times \sqrt{b^5c^4}\)
----
\(\dfrac{\sqrt{12}}{\sqrt{3}}\)
----
\(\sqrt{x^2y^{13}} \div \sqrt{y^5}\)
----
\(\left( \dfrac{1}{a} - \dfrac{1}{b} \right)^{-1}\)
----
\(\dfrac{b-a}{a^{\frac{1}{2}} - b^{\frac{1}{2}}}\)
----
\[\frac{5x-16}{\sqrt{x}}\]
----
\(\frac{\sqrt{x}}{\sqrt{x}}\)
----
\(\frac{\sqrt{x}}{\sqrt{x}}=1\)
----
\[\frac{5x - 16}{\sqrt{x}} \times \frac{\sqrt{x}}{\sqrt{x}} = \frac{\sqrt{x}(5x - 16)}{\sqrt{x} \times \sqrt{x}}\]
----
\[\begin{align}  			
		    &= \frac{\sqrt{x}(5x - 16)}{\left( \sqrt{x}\right)^2} \\
		    &= \frac{\sqrt{x}(5x - 16)}{x}
		    \end{align}\]
----
\[\frac{y-25}{\sqrt{y}+5}\]
----
\(\frac{\sqrt{y}-5}{\sqrt{y}-5}\)
----
\[\frac{y-25}{\sqrt{y}+5} \times \frac{\sqrt{y}-5}{\sqrt{y}-5}\]
----
\[\begin{align}  			
		    &= \frac{(y-25)(\sqrt{y}-5)}{(\sqrt{y}+5)(\sqrt{y}-5)} \\
		    &= \frac{(y-25)(\sqrt{y}-5)}{(\sqrt{y})^2 -25} \\
		    &= \frac{(y-25)(\sqrt{y}-5)}{y-25} \\
		    &= \sqrt{y}-5
		    \end{align}\]
----
\(\dfrac{10}{\sqrt{5}}\)
----
\(\dfrac{3}{\sqrt{6}}\)
----
\(\dfrac{2}{\sqrt{3}} \div \dfrac{\sqrt{2}}{3}\)
----
\(\dfrac{3}{\sqrt{5}-1}\)
----
\(\dfrac{x}{\sqrt{y}}\)
----
\(\dfrac{\sqrt{3} + \sqrt{7}}{\sqrt{2}}\)
----
\(\dfrac{3\sqrt{p} - 4}{\sqrt{p}}\)
----
\(\dfrac{t-4}{\sqrt{t} + 2}\)
----
\(\left( 1 + \sqrt{m} \right)^{-1}\)
----
\(a \left( \sqrt{a} \div \sqrt{b} \right)^{-1}\)
----
\(x\)
----
\[\begin{align}  			
		  \left( x^{\frac{4}{3}} \right)^{\frac{3}{4}} &= \left( 3^4 \right)^{\frac{3}{4}} \\
		  x  &= 3^3 \\
		  x  &= 27
		  \end{align}\]
----
\(z\)
----
\(z - 4\sqrt{z} + 3 = 0\)
----
\[\begin{align} 
		  z - 4\sqrt{z} + 3   &= 0 \\ 			
		  z - 4z^{\frac{1}{2}} + 3   &= 0 \\
		  (z^{\frac{1}{2}}-3)(z^{\frac{1}{2}}-1)   &= 0
		  \end{align}\]
----
\(a \times b = 0\)
----
\(a = 0\)
----
\(b = 0\)
----
\[\therefore (z^{\frac{1}{2}}-3) = 0 \text{ or } (z^{\frac{1}{2}}-1) = 0\]
----
\[\begin{align}  			
		  z^{\frac{1}{2}}-3 &= 0 \\
		  z^{\frac{1}{2}}   &= 3 \\
		  \left( z^{\frac{1}{2}} \right)^2 &= 3^2 \\
		  z &= 9 
		  \end{align}\]
----
\[\begin{align}  			
		  z^{\frac{1}{2}}-1 &= 0 \\
		  z^{\frac{1}{2}}   &= 1 \\
		  \left( z^{\frac{1}{2}} \right)^2 &= 1^2 \\
		  z &= 1
		  \end{align}\]
----
\(z=9\)
----
\[\begin{align}  			
		  \text{LHS}&= z - 4\sqrt{z} + 3   \\
		  &= 9 - 4\sqrt{9} + 3    \\
		  &= 12 - 12   \\
		  &=  0   \\
		  &=\text{RHS }
		  \end{align}\]
----
\(z=1\)
----
\[\begin{align}  			
		  \text{LHS} &= z - 4\sqrt{z} + 3   \\
		  &= 1 - 4\sqrt{1} + 3  \\
		  &= 4-4  \\
		  &=  0   \\
		  &= \text{RHS }
		  \end{align}\]
----
\(z - 4\sqrt{z} + 3 = 0\)
----
\(z = 9\)
----
\(z = 1\)
----
\(p\)
----
\(\sqrt{p-2} - 3 = 0\)
----
\[\sqrt{p-2} = 3\]
----
\[\begin{align}  			
		  \left( \sqrt{p-2} \right)^2 &= 3^2 \\
		  p-2 &= 9 \\
		  p &= 11
		  \end{align}\]
----
\(p=11\)
----
\[\begin{align}  			
		  \text{LHS} &=\sqrt{p-2} - 3   \\
		  &=\sqrt{11-2} - 3  \\
		  &=\sqrt{9} - 3  \\
		  &= 3-3   \\
		  &= 0   \\
		  &= \text{RHS } 
		  \end{align}\]
----
\(\sqrt{p-2} - 3 = 0\)
----
\(p = 11\)
----
\(2^{x+1} -32 = 0\)
----
\[\begin{aligned}
2^{x+1}-32&=0 \\
2^{x+1}&=32 \\
2^{x+1}&=2^5 \\
\therefore x+1&=5 \\
x&=4
\end{aligned}\]
----
\(x = 4\)
----
\(p = 3\)
----
\(2y^{\frac{1}{2}} - 3y^{\frac{1}{4}} + 1 = 0\)
----
\[\begin{aligned}
2y^{\frac{1}{2}}-3y^{\frac{1}{4}}+1&=0 \\
\left ( 2y^\frac{1}{4} -1\right )\left ( y^\frac{1}{4} -1\right )&=0 \\
\text{Therefore } 2y^{\frac{1}{4}}-1&=0  \\
y^{\frac{1}{4}}-\frac{1}{2}&=0 \\
y^\frac{1}{4}&=\frac{1}{2} \\
y^\frac{4}{4} &= \left ( \frac{1}{2} \right )^4 \\
\therefore y&=\frac{1}{16} \\
\text{or} \\
y^\frac{1}{4}-1&=0 \\
y^\frac{1}{4}&=1 \\
\therefore y&=1
\end{aligned}\]
----
\(y = 1\)
----
\(t-1 = \sqrt{7-t}\)
----
\[\begin{aligned}
t-1 &= \sqrt{7-t} \\
\left ( t-1 \right )^2 &= \left ( \sqrt{7-t} \right )^2 \\
t^2 - 2t + 1 &=  7-t  \\
t^2 - t - 6 &=0 \\
(t - 3)(t + 2)&= 0 \\
\therefore  t =3  &\text{ or } t = - 2 \\
\text{Check RHS for } t = 3:  &= \sqrt{7-3} \\
   						&= \sqrt{4} \\
   						&= 2 \\
   						&= \text{LHS} \therefore \text{ valid solution }\\
\text{Check RHS for } t = -2:  &= \sqrt{7-(-2)} \\
   						&= \sqrt{9} \\
   						&= 3 \\
   						&\ne \text{LHS} \therefore \text{ not valid solution } 
\end{aligned}\]
----
\(t = 3\)
----
\(2z - 7\sqrt{z} + 3 = 0\)
----
\[\begin{aligned}
2z-7z^\frac{1}{2}+3&=0 \\
\left ( z^\frac{1}{2}-3 \right )\left ( 2z^\frac{1}{2}-1 \right )&=0 \\
\text{Therefore } z^\frac{1}{2}-3&=0 \\
\left ( z^\frac{1}{2} \right )^2 &=3^2 \\
\therefore z&=9 \\
\text{or}\\
2z^\frac{1}{2}-1&=0 \\
\left ( z^\frac{1}{2} \right )^2&=\left ( \frac{1}{2} \right )^2 \\
\therefore z&=\frac{1}{4}
\end{aligned}\]
----
\(z = 9\)
----
\(z = \frac{1}{4}\)
----
\(x^{\frac{1}{3}}(x^{\frac{1}{3}} + 1) = 6\)
----
\[\begin{aligned}
x^\frac{1}{3}\left ( x^\frac{1}{3}+1 \right )&=6 \\
x^\frac{2}{3}+x^\frac{1}{3} &=6 \\
\left ( x^\frac{1}{3}-2 \right )\left ( x^\frac{1}{3}+3 \right ) &= 0 \\
\text{Therefore } x^\frac{1}{3}-2 &= 0 \\
x^\frac{1}{3} &= 2 \\
\therefore x&=8 \\
\text{or} \\
x^\frac{1}{3}+3 &= 0 \\
x^\frac{1}{3} &=-3 \\
\therefore x &= -27
\end{aligned}\]
----
\(x = 8\)
----
\(x = -27\)
----
\(2^{4n} - \dfrac{1}{\sqrt[4]{16}} = 0\)
----
\[\begin{aligned}
2^{4n}-\dfrac{1}{\sqrt[4]{16}}&=0\\
2^{4n}&=\dfrac{1}{16^\frac{1}{4}} \\
2^{4n}&=\dfrac{1}{\left (2^4  \right )^\frac{1}{4}} \\
2^{4n}&=\frac{1}{2} \\
2^{4n}&=2^{-1} \\
\therefore 4n&=-1 \\
\therefore n &=-\frac{1}{4}
\end{aligned}\]
----
\(n = -\frac{1}{4}\)
----
\(\sqrt{31 -10d} = 4 - d\)
----
\[\begin{aligned}
\sqrt{31-10d}&=d-4 \\
\left (\sqrt{31-10d}  \right )^2&=\left (d-4  \right )^2 \\
31-10d&=d^2-8d+16 \\
0&=d^2+2d-15 \\
0&=\left ( d-3 \right )\left ( d+5 \right ) \\ 
\text{Therefore } d&= 3 \\
\text{or} \\
d&= -5 \\
\text{Check LHS for } d = 3:  &= \sqrt{31 - 30} \\
   						&= \sqrt{1} \\
   						&= 1 \\
   						&= \text{RHS} \therefore \text{ valid solution }\\
\text{Check LHS for } t = -5:  &= \sqrt{31-(-50)} \\
   						&= \sqrt{81} \\
   						&= 9 \\
   						&= \text{RHS} \therefore \text{ valid solution }
\end{aligned}\]
----
\(d = 3\)
----
\(d = -5\)
----
\(y - 10\sqrt{y} + 9 = 0\)
----
\[\begin{aligned}
y - 10\sqrt{y} + 9 &= 0  \\
y - 10y^{\frac{1}{2}} + 9 &= 0  \\
\left( y^{\frac{1}{2}} - 1  \right ) \left( y^{\frac{1}{2}} -9 \right )  &= 0 \\
\text{Therefore } y^{\frac{1}{2}} - 1  &= 0 \\
y^{\frac{1}{2}}  &= 1 \\
\left( y^{\frac{1}{2}} \right )^2 &= (1)^2 \\
\therefore y  &= 1 \\
\text{or} \\
y^{\frac{1}{2}} - 9 & = 0  \\
y^{\frac{1}{2}}  &= 9 \\
\left( y^{\frac{1}{2}} \right )^2 &= (9)^2 \\
\therefore y  &= 81 
\end{aligned}\]
----
\(y = 1\)
----
\(y = 81\)
----
\(f = 2 + \sqrt{19 - 2f}\)
----
\[\begin{aligned}
f &= 2 + \sqrt{19 - 2f} \\
f - 2 &=   \sqrt{19 - 2f} \\
\left( f - 2 \right )^2 &=   \left( \sqrt{19 - 2f} \right )^2  \\
f^2 - 4f + 4 &= 19 - 2f  \\
 f^2 - 2f -15  &= 0 \\
(f - 5)(f + 3)  &= 0 \\
\text{Therefore } f - 5  &= 0 \\
\therefore f  &= 5 \\
\text{or} \\
f + 3  & = 0  \\
\therefore f  &= -3 \\
\text{Check RHS for } f = 5:  &= 2 + \sqrt{19 - 10} \\
   						&= 2 + \sqrt{9} \\
   						&= 5 \\
   						&= \text{LHS} \therefore \text{ valid solution }\\
\text{Check RHS for } f = -3:  &= 2 + \sqrt{19 + 6}\\
   						&= 2 + \sqrt{25} \\
   						&= 7 \\
   						&\ne \text{LHS} \therefore \text{ not valid solution }
\end{aligned}\]
----
\(f = 5\)
----
\[\text{final population} = \text{initial population} \times (1 + \text{growth percentage})^{\text{time period in hours}}\]
----
\(n =\)
----
\[\text{final population} = \text{initial population} \times (1+\text{growth percentage})^{\text{time period in months}}\]
----
\(n =\)
----
\(3; 6; 12; 24; 48; \ldots\)
----
\(ℕ\)
----
\(\{1; \; 2; \; 3; \; \ldots\}\)
----
\(ℕ_0\)
----
\(\{0; \; 1; \; 2; \; 3; \; \ldots\}\)
----
\(ℤ\)
----
\(\{\ldots; \; -3; \; -2; \; -1; \; 0; \; 1; \; 2; \; 3; \; \ldots\}\)
----
\(ℚ\)
----
\(\frac{a}{b}\)
----
\(a\)
----
\(b\)
----
\(b\ne 0\)
----
\(ℚ'\)
----
\(ℝ\)
----
\(ℝ'\)
----
\({a}^{n}=a\times a\times a\times \cdots \times a\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\left(n\phantom{\rule{3.33333pt}{0ex}}\text{times}\right)\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\phantom{\rule{3.33333pt}{0ex}}\left(a\in ℝ,n\in ℕ\right)\)
----
\({a}^{0}=1\phantom{\rule{28.45274pt}{0ex}}\)
----
\(a \ne 0\)
----
\(0^0\)
----
\({a}^{-n}=\frac{1}{{a}^{n}}\phantom{\rule{14.22636pt}{0ex}}\)
----
\(a \ne 0\)
----
\(\dfrac{1}{0}\)
----
\({a}^{m} \times {a}^{n}={a}^{m+n}\)
----
\(\dfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}\)
----
\({\left(ab\right)}^{n}={a}^{n}{b}^{n}\)
----
\({\left(\frac{a}{b}\right)}^{n}=\dfrac{{a}^{n}}{{b}^{n}}\)
----
\({\left({a}^{m}\right)}^{n}={a}^{mn}\)
----
\(a > 0\)
----
\(b > 0\)
----
\(m, n \in ℤ\)
----
\(r^n = a\)
----
\(r = \sqrt[n]{a} \quad (n \geq 2)\)
----
\(a^{\frac{1}{n}} = \sqrt[n]{a}\)
----
\(a^{-\frac{1}{n}} = (a^{-1})^{\frac{1}{n}} = \sqrt[n]{\dfrac{1}{a}}\)
----
\(a^{\frac{m}{n}} = (a^{m})^{\frac{1}{n}} = \sqrt[n]{a^m}\)
----
\(a > 0\)
----
\(r > 0\)
----
\(m,n \in ℤ\)
----
\(n \ne 0\)
----
\(\sqrt[n]{a}\sqrt[n]{b} = \sqrt[n]{ab}\)
----
\(\sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}\)
----
\(\sqrt[m]{\sqrt[n]{a}} = \sqrt[mn]{a}\)
----
\({8}^{-\frac{2}{3}}\)
----
\(\sqrt{16}+{8}^{-\frac{2}{3}}\)
----
\({\left({x}^{3}\right)}^{\frac{4}{3}}\)
----
\({\left({s}^{2}\right)}^{\frac{1}{2}}\)
----
\({\left({m}^{5}\right)}^{\frac{5}{3}}\)
----
\({\left(-{m}^{2}\right)}^{\frac{4}{3}}\)
----
\(-{\left({m}^{2}\right)}^{\frac{4}{3}}\)
----
\({\left(3{y}^{\frac{4}{3}}\right)}^{4}\)
----
\(\dfrac{3{a}^{-2}{b}^{15}{c}^{-5}}{{\left({a}^{-4}{b}^{3}c\right)}^{\frac{-5}{2}}}\)
----
\({\left(9{a}^{6}{b}^{4}\right)}^{\frac{1}{2}}\)
----
\({\left({a}^{\frac{3}{2}}{b}^{\frac{3}{4}}\right)}^{16}\)
----
\({x}^{3}\sqrt{x}\)
----
\(\sqrt[3]{{x}^{4}{b}^{5}}\)
----
\(x\)
----
\(\dfrac{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}}{x^2}\)
----
\(\left(\sqrt{x}-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{2}\right)\)
----
\(\dfrac{10}{\sqrt{x}-\frac{1}{x}}\)
----
\(\dfrac{3}{2\sqrt{x}}+\sqrt{x}\)
----
\(\sqrt{72}\)
----
\(\sqrt{45}+\sqrt{80}\)
----
\(\dfrac{\sqrt{48}}{\sqrt{12}}\)
----
\(\dfrac{\sqrt{18}÷\sqrt{72}}{\sqrt{8}}\)
----
\(\dfrac{4}{\left(\sqrt{8}÷\sqrt{2}\right)}\)
----
\(\dfrac{16}{\left(\sqrt{20}÷\sqrt{12}\right)}\)
----
\({\left(2+\sqrt{2}\right)}^{2}\)
----
\(\left(2+\sqrt{2}\right)\left(1+\sqrt{8}\right)\)
----
\(\left(1+\sqrt{3}\right)\left(1+\sqrt{8}+\sqrt{3}\right)\)
----
\(\sqrt{5}\left(\sqrt{45}+2\sqrt{80}\right)\)
----
\(\dfrac{\sqrt{98}-\sqrt{8}}{\sqrt{50}}\)
----
\(\sqrt{98{x}^{6}}+\sqrt{128{x}^{6}}\)
----
\(\dfrac{\sqrt{5}+2}{\sqrt{5}}\)
----
\(\dfrac{y-4}{\sqrt{y}-2}\)
----
\(\dfrac{2x-20}{\sqrt{x}-\sqrt{10}}\)
----
\({\left(2-\dfrac{\sqrt{7}}{2}\right)}^{\frac{1}{2}}\phantom{\rule{0.222222em}{0ex}}\times \phantom{\rule{0.222222em}{0ex}}{\left(2+\dfrac{\sqrt{7}}{2}\right)}^{\frac{1}{2}}\)
----
\(\sqrt{\dfrac{8}{3}}+5\sqrt{\dfrac{5}{3}}-\sqrt{\dfrac{1}{6}}=\dfrac{10\sqrt{15}+3\sqrt{6}}{6}\)
----
\[\begin{aligned}
 \text{LHS }  &= \sqrt{\dfrac{8}{3}} + 5 \sqrt{\dfrac{5}{3}} - \sqrt{\dfrac{1}{6}}  \\
				&=  \sqrt{\frac{8}{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} + 5 \sqrt{\dfrac{5}{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} - \sqrt{\dfrac{1}{6}} \times \dfrac{\sqrt{6}}{\sqrt{6}} \\
				&=   \dfrac{\sqrt{8}\sqrt{3}}{3}  + 5\frac{\sqrt{5}\sqrt{3}}{3} -  \frac{\sqrt{6}}{6} \\
				&=   \dfrac{ 2\sqrt{24}}{6}  + \frac{10\sqrt{15}}{6} -  \frac{\sqrt{6}}{6} \\
				&=   \dfrac{ 4\sqrt{6}  + 10\sqrt{15} - \sqrt{6}}{6} \\
				&=  \dfrac{ 10\sqrt{15} + 3\sqrt{6}}{6} \\
				&= \text{RHS }
\end{aligned}\]
----
\[{3}^{-\frac{1}{2}}\left[\sqrt{12}+\sqrt[3]{\left(3\sqrt{3}\right)}\right]\]
----
\(-3\sqrt{6}\times -2\sqrt{24}=-\sqrt{18}\times ...\)
----
\(\)
----
\(3^{x - 1} - 27 = 0\)
----
\(8^x - \frac{1}{\sqrt[3]{8}} = 0\)
----
\(27(4^x) = (64)3^x\)
----
\(\sqrt{2x - 5} = 2 - x\)
----
\(2x^{\frac{2}{3}} + 3x^{\frac{1}{3}} - 2 = 0\)
----
\[\begin{aligned}
	2x^{\frac{2}{3}} + 3x^{\frac{1}{3}} - 2 &= 0 \\
	\left( 2x^{\frac{1}{3}} - 1  \right) \left( x^{\frac{1}{3}}  + 2  \right) &= 0 \\
		\therefore 2x^{\frac{1}{3}} - 1 &= 0 \\
		2x^{\frac{1}{3}}	&= 1 \\
		x^{\frac{1}{3}}	&= \frac{1}{2} \\
		\left( x^{\frac{1}{3}} \right)^3	&= \left( \frac{1}{2} \right)^3 \\
		\therefore x	&= \frac{1}{8}  \\
		\text{ or } &  \\ 
 	 x^{\frac{1}{3}}  + 2  &=  0 \\ 
 	 x^{\frac{1}{3}}   &=  -2 \\ 
 	\left( x^{\frac{1}{3}} \right)^3  &= \left( -2 \right)^3 \\ 
 	 x  &= -8 \\ 
 \text{Therefore } x = \frac{1}{8} &\text{ or } x = -8  
\end{aligned}\]
----
\(x = \frac{1}{8} \text{ or } x = -8\)
----
\(\sqrt{\dfrac{3^{x+1} - 3^x}{3^{x-1}} + 3}\)
----
\[\begin{aligned}
 \text{LHS }  &= \sqrt{\dfrac{3^{x+1} - 3^x}{3^{x-1}} + 3} \\
				&=  \sqrt{\dfrac{3^{x+1}}{3^{x-1}} - \dfrac{3^{x}}{3^{x-1}}+ 3} \\
				&=  \sqrt{3^{x+1-x+1} -  3^{x-x+1} + 3} \\
				&=  \sqrt{3^{2} -  3^{1} + 3} \\
				&=  \sqrt{3^{2}} \\
				&=  3 \\
				&=   \text{RHS }
\end{aligned}\]
----
\(\sqrt{\dfrac{3^{x+1} - 3^x}{3^{x-1}} + 3} = \left( \dfrac{1}{3} \right)^{x-2}\)
----
\[\begin{aligned}
 \sqrt{\dfrac{3^{x+1} - 3^x}{3^{x-1}} + 3} &=  \left( \dfrac{1}{3} \right)^{x-2} \\
 3  &=  \left( \frac{1}{3} \right)^{x-2} \\
 3  &=  \left( 3\right)^{-x+2} \\
		1		&=  -x + 2\\
		x		&=  1
\end{aligned}\]
----
\(x = 1\)
----
\(\pm\)
----
\(x = -b \pm \frac{b^2 – 4ac}{2a}\)
----
\(+, -, \times, \div, x^n\)
----
\(>, <, \geq, \leq\)
----
\(x\)
----
\(f(x)=0\)
----
\[\begin{align}
	    2x^2-5x &= 12 \\
	    a(a-3)-10 &= 0 \\
	    \frac{3b}{b+2} + 1 &= \frac{4}{b+1}
	    \end{align}\]
----
\(\begin{aligned}
  y &= (x-2)(x+2) \\
  &= x^2 - 4
\end{aligned}\)
----
\(x = -2\)
----
\(x = 2\)
----
\(y = x^2\)
----
\(x = 0\)
----
\(y = x^2 + x + 1\)
----
\(ax^2+bx+c=0\)
----
\(6 \times 0 = \text{?}\)
----
\(-25 \times 0 = \text{?}\)
----
\(7 \times \text{?} = 0\)
----
\(6 \times m = 0\)
----
\(32 \times x \times 2 = 0\)
----
\(11(z-3) = 0\)
----
\((k+3)(k-4) = 0\)
----
\(a\times b=0\)
----
\(a=0\)
----
\(b=0\)
----
\(a{x}^{2}+bx+c=0\)
----
\(a{x}^{2}+bx+c=0\)
----
\(a\)
----
\(b\)
----
\(c\)
----
\(a{x}^{2}+bx+c=0\)
----
\(\left(rx+s\right)\left(ux+v\right)=0\)
----
\[\begin{array}{r@{\;}l@{\qquad\qquad}r@{\;}l}
      (rx+s) &= 0             & (ux+v) &= 0 \\[8pt]
\text{So } x &= -\dfrac{s}{r} & \text{So } x &= -\dfrac{v}{u}
\end{array}\]
----
\(x\)
----
\(\phantom{\rule{3.33333pt}{0ex}}x\left(x-3\right)=10\)
----
\(a{x}^{2}+bx+c=0\)
----
\({x}^{2}-3x-10=0\)
----
\(\left(x+2\right)\left(x-5\right)=0\)
----
\[\begin{align}
		    x+2 &= 0 \\
		    x &= -2 
		    \end{align}\]
----
\[\begin{align}
		    x-5 &= 0 \\
		    x &= 5 
		    \end{align}\]
----
\(x=-2\)
----
\(x=5\)
----
\(x=-2\)
----
\(x=5\)
----
\(2{x}^{2}-5x-12=0\)
----
\(a{x}^{2}+bx+c=0\)
----
\(2 \text{ and } 12\)
----
\(2 \times 1 \text{ and } 3 \times 4\)
----
\[(2x+3)(x-4) = 0\]
----
\[\begin{align}
		    2x+3 &= 0 \\
		    x &= - \frac{3}{2}
		    \end{align}\]
----
\[\begin{align}
		    x-4 &= 0 \\
		    x &= 4 
		    \end{align}\]
----
\(x=-\frac{3}{2}\)
----
\(x=4\)
----
\(y\)
----
\(y^2-7 = 0\)
----
\[\left( \sqrt{7} \right)^2 = 7\]
----
\[y^2 - (\sqrt{7})^2 = 0\]
----
\[\begin{align}
		    (y - \sqrt{7})(y + \sqrt{7})  &= 0 \\
		    \text{Therefore } y = \sqrt{7} &\text{ or }  y = -\sqrt{7}
		    \end{align}\]
----
\(f(y) = y^2-7\)
----
\(y = \pm \sqrt{7}\)
----
\(b\)
----
\[\frac{3b}{b+2} + 1 = \frac{4}{b+1}\]
----
\(b\)
----
\(\left(b+2\right)\left(b+1\right)\)
----
\[\begin{align}
		\frac{3b(b+2)(b+1)}{b+2} + (b+2)(b+1) &=  \frac{4(b+2)(b+1)}{b+1} \\
		3b(b+1) + (b+2)(b+1) &= 4(b+2) \\
		3b^2 + 3b + b^2+3b+2 &= 4b+8 \\
		4b^2 + 2b - 6 &= 0 \\
		2b^2 + b - 3 &= 0 
		\end{align}\]
----
\[\begin{align}
		(2b + 3)(b - 1)&=  0\\
		2b + 3 =0 & \text{ or } b-1 = 0 \\
		b = -\frac{3}{2} & \text{ or } b = 1
		\end{align}\]
----
\(b=-1\frac{1}{2}\)
----
\(b = 1\)
----
\(m\)
----
\(m + 2 = \sqrt{7 + 2m}\)
----
\[(m + 2)^2 = \left( \sqrt{7 + 2m} \right)^2\]
----
\[\begin{align}
								(m + 2)^2 &= \left( \sqrt{7 + 2m} \right)^2 \\
								m^2 + 4m + 4 &= 7 + 2m \\
								m^2 + 2m - 3 &= 0
								\end{align}\]
----
\(m\)
----
\[\begin{align}
		  (m - 1)(m + 3) &= 0 \\
		  \text{Therefore } m = 1 &\text{ or } m = -3
		  \end{align}\]
----
\(m=1\)
----
\[\begin{align}
		  \text{RHS } &= \sqrt{7 + 2(1)} \\
		  &= \sqrt{9} \\
		  &= 3 \\
		  \text{LHS } &= 1 + 2 \\
		  &= 3 \\
		  \text{LHS } &= \text{RHS }
		  \end{align}\]
----
\(m = 1\)
----
\(m=-3\)
----
\[\begin{align}
		  \text{RHS } &= \sqrt{7 + 2(-3)} \\
		  &= \sqrt{1} \\
		  &= 1 \\
		  \text{LHS } &= -3 + 2 \\
		  &= -1 \\
		  \text{LHS } & \ne \text{RHS }
		  \end{align}\]
----
\(m = -3\)
----
\(m = 1\)
----
\(7t^2+14t=0\)
----
\[\begin{align}
  7t^2 + 14t &= 0 \\
  7t(t+2) &= 0 \\
  t(t+2) &= 0 \\
  t = 0 &\text{ or } t = -2
\end{align}\]
----
\(t=0\)
----
\(t=-2\)
----
\(12{y}^{2}+24y+12=0\)
----
\[\begin{align}
		  12y^2 + 24y + 12 &=  0\\
		  12(y^2 + 2y+1)&= 0\\
		  y^2 + 2y+1&= 0\\
		  (y+1)(y+1)&= 0\\
		  y &=  -1
		  \end{align}\]
----
\(y=-1\)
----
\(16s^2 = 400\)
----
\[\begin{align}
		  16s^2 - 400 &= 0\\
		  16(s^2 - 25)&= 0\\
		  (s-5)(s+5) &= 0\\
		  s &= \pm 5
		  \end{align}\]
----
\(s=\pm 5\)
----
\({y}^{2}-5y+6=0\)
----
\[\begin{align}
		  y^2 - 5y + 6 &= 0\\
		  (y-3)(y-2) &= 0\\
		  y = 3 &\text{ or } y=2
		  \end{align}\]
----
\(y=3\)
----
\(y=2\)
----
\({y}^{2}+5y-36=0\)
----
\[\begin{align}
		  y^2 + 5y - 36 &= 0\\
		  (y-4)(y+9) &= 0\\
		  y = 4 &\text{ or } y = -9
		  \end{align}\]
----
\(y=4\)
----
\(y=-9\)
----
\(4+p = \sqrt{p+6}\)
----
\[\begin{align}
		  4 + p & = \sqrt{p + 6}\\
		  (4 + p)^{2} &= p + 6\\
		  16 + 8p + p^{2} & = p + 6 \\
		  p^{2} + 7p + 10 & = 0 \\
		  (p + 5)(p + 2) & = 0 \\
		  p = -5 &\text{ or } p = -2
		  \end{align}\]
----
\(p=-5\)
----
\[\begin{align}
		    \text{RHS } &= \sqrt{-5 +6} \\
		    &= \sqrt{1} \\ 
		    &= 1 \\ 
		    \text{LHS } &= 4 + (-5)  \\ 
		    &= -1 \\
		    \therefore \text{LHS} &\ne \text{RHS}
		    \end{align}\]
----
\(p = -5\)
----
\(p = -2\)
----
\[\begin{align}
		    \text{RHS } &= \sqrt{-2 + 6}  \\ 
		    &= \sqrt{4} \\ 
			& = 2 \\
		    \text{LHS } &= 4 + (-2)  \\ 
		    &= 2 \\
		    \therefore \text{LHS} &= \text{RHS}
		    \end{align}\]
----
\(p = -2\)
----
\(p = -2\)
----
\(p=-2\)
----
\(-{y}^{2}-11y-24=0\)
----
\[\begin{align}
		  -y^2 - 11y - 24 &= 0\\
		  y^2 + 11y +24 &= 0\\
		  (y+3)(y+8) &= 0\\
		  y = -3 &\text{ or } y = -8
		  \end{align}\]
----
\(y=-3\)
----
\(y=-8\)
----
\(13y-42={y}^{2}\)
----
\[\begin{align}
		  13y - 42 &= y^2\\
		  y^2 - 13y + 42 &= 0\\
		  (y-6)(y-7) &=0\\
		  y = 6 &\text{ or } y=7
		  \end{align}\]
----
\(y=6\)
----
\(y=7\)
----
\((x-1)(x+10) = -24\)
----
\[\begin{align}
		  (x-1)(x+10) &= -24 \\
		  x^2 + 9x - 10 &= -24 \\
		  x^2 + 9x + 14 &= 0 \\
		  (x+7)(x+2) &= 0\\
		  x = -7 &\text{ or } x=-2
		  \end{align}\]
----
\(x=-7\)
----
\(x=-2\)
----
\(y^2 - 5ky + 4k^2 = 0\)
----
\[\begin{align}
		  y^2 - 5ky + 4k^2 &= 0\\
		  (y-4k)(y-k) &= 0\\
		  y = 4k &\text{ or } y=k 
		  \end{align}\]
----
\(y=4k\)
----
\(y=k\)
----
\(2{y}^{2}-61=101\)
----
\[\begin{align}
		  2y^2 - 61 &=101 \\
		  2y^2 - 162 &=0\\
		  2(y^2-81)&=0\\
		  y^2-81 &=0\\
		  (y-9)(y+9) &= 0\\
		  y= 9 &\text{ or } y= -9 
		  \end{align}\]
----
\(y=9\)
----
\(y=-9\)
----
\(2{y}^{2}-10=0\)
----
\[\begin{align}
		  2y^2 - 10 &= 0\\
		  2(y^2-5)&=0\\
		  y^2-5&=0\\
		  (y-\sqrt{5})(y+\sqrt{5}) &= 0\\
		  y = \sqrt{5} &\text{ or } y=-\sqrt{5}
		  \end{align}\]
----
\(y=\pm \sqrt{5}\)
----
\(-8 + h^2 = 28\)
----
\[\begin{align}
		  -8 + h^2 &= 28 \\
		  h^2 - 36 &= 0\\
		  (h-6)(h+6) &= 0\\
		  h &= \pm 6
		  \end{align}\]
----
\(h = \pm 6\)
----
\({y}^{2}-4=10\)
----
\[\begin{align}
		  y^2 - 4 &=10\\
		  y^2 - 14 &= 0\\
		  (y - \sqrt{14})(y+\sqrt{14}) &= 0\\
		  y = \sqrt{14} &\text{ or } y=-\sqrt{14}
		  \end{align}\]
----
\(y=\pm \sqrt{14}\)
----
\(\sqrt{5-2p} - 4 = \frac{1}{2}p\)
----
\[\begin{align}
		  \sqrt{5-2p} - 4  &= \frac{1}{2}p \\
		  \sqrt{5-2p}  &= \frac{1}{2}p + 4 \\
		  \left( \sqrt{5-2p} \right) ^2  &= \left( \frac{1}{2}p + 4 \right)^2 \\
		  5-2p   &=   \frac{p^2}{4} + 4p +16   \\
		  0   &=   \frac{p^2}{4} + 6p + 11   \\
		  0   &=   p^2 + 24p + 44   \\
		  0   &=  (p + 2)(p + 22)   \\
		  p = -2 &\text{ or } p=-22
		  \end{align}\]
----
\(p=-2\)
----
\[\begin{align}
		    \text{LHS } &= \sqrt{5 -2(-2)} - 4 \\
		    &= \sqrt{9} - 4 \\ 
		    &= -1 \\ 
		    \text{RHS } &= \frac{1}{2}(-2)  \\ 
		    &= -1 \\
		    \therefore \text{LHS} &= \text{RHS}
		    \end{align}\]
----
\(p = -2\)
----
\(p = -22\)
----
\[\begin{align}
		    \text{LHS } &= \sqrt{5 -2(-22)} - 4  \\ 
		    &= \sqrt{49} - 4 \\ 
		    &= 3 \\ 
		    \text{RHS } &= \frac{1}{2}(-22)  \\ 
		    &= -11 \\
		    \therefore \text{LHS} &\ne \text{RHS}
		    \end{align}\]
----
\(p = -22\)
----
\(p = -2\)
----
\({y}^{2}+28=100\)
----
\[\begin{align}
		  y^2 + 28 &= 100\\
		  y^2-72 &= 0\\
		  (y-\sqrt{72})(y+\sqrt{72}) &=0\\
		  y &= \pm \sqrt{72}\\
		  &= \pm \sqrt{2 \times 36}\\
		  &= \pm 6\sqrt{2}
		  \end{align}\]
----
\(y=\pm 6\sqrt{2}\)
----
\(f\left(2f+1\right)=15\)
----
\[\begin{align}
		  f(2f + 1) &=15\\
		  2f^2 + f - 15 &= 0\\
		  (2f - 5)(f+3) &= 0\\
		  f =\frac{5}{2} &\text{ or } f = -3
		  \end{align}\]
----
\(f= \frac{5}{2}\)
----
\(f=-3\)
----
\(2x = \sqrt{21x - 5}\)
----
\[\begin{align}
		   2x  &= \sqrt{21x - 5} \\
		  (2x)^{2}  &= \left(\sqrt{21x - 5} \right)^{2} \\
		  4x^{2}  &= 21x - 5 \\
		  4x^{2} - 21x + 5  &= 0 \\
		  (4x - 1)(x - 5) & = 0 \\
		  x = \frac{1}{4} &\text{ or } x= 5
		  \end{align}\]
----
\(x= \frac{1}{4}\)
----
\[\begin{align}
		    \text{LHS } &= \sqrt{21\left(\frac{1}{4} \right) - 5} \\
		    &= \sqrt{\frac{1}{4}} \\ 
		    &= \frac{1}{2} \\ 
		    \text{RHS } &= \frac{1}{4}(2)  \\ 
		    &= \frac{1}{2} \\
		    \therefore \text{LHS} &= \text{RHS}
		    \end{align}\]
----
\(x = \frac{1}{4}\)
----
\(x = 5\)
----
\[\begin{align}
		    \text{LHS } &= \sqrt{21(5) - 5}  \\ 
		    &= \sqrt{100} \\ 
		    &= 10 \\ 
		    \text{RHS } &= 2(10)  \\ 
		    &= 20 \\
		    \therefore \text{LHS} &\ne \text{RHS}
		    \end{align}\]
----
\(x = 5\)
----
\(x = \frac{1}{4}\)
----
\(\dfrac{5y}{y-2}+\dfrac{3}{y}+2=\dfrac{-6}{{y}^{2}-2y}\)
----
\(y \ne 0\)
----
\(y \ne 2\)
----
\[\dfrac{5y}{y - 2}+\dfrac{3}{y}+ 2 = d\frac{-6}{y^2 - 2y}\]
----
\(y(y-2)\)
----
\[\begin{align}
		    y(5y) + 3(y-2) + 2y(y-2) &= -6\\
		    5y^2 + 3y - 6 + 2y^2 - 4y &= -6\\
		    7y^2 - y &= 0\\
		    y(7y-1) &= 0\\
		    y = 0 &\text{ or } y = \frac{1}{7}
		    \end{align}\]
----
\(y \ne 0\)
----
\(y = \frac{1}{7}\)
----
\(\dfrac{x+9}{x^2-9} + \dfrac{1}{x+3} = \dfrac{2}{x-3}\)
----
\(x \ne \pm 3\)
----
\[\dfrac{x+9}{(x+3)(x-3)} +  \dfrac{1}{x+3} \dfrac{2}{x-3}\]
----
\((x+3)(x-3)\)
----
\[\begin{align}
		    x+9 + x-3 &= 2(x+3) \\
		    2x+6 &= 2x+6 \\
		    0x &= 0
		    \end{align}\]
----
\(x \in \mathbb{R}\)
----
\(x \ne \pm 3\)
----
\(\dfrac{y-2}{y+1}=\dfrac{2y+1}{y-7}\)
----
\(y \ne -1\)
----
\(y \ne 7\)
----
\[\begin{align}
		    \dfrac{y-2}{y+1} &= \dfrac{2y+1}{y-7}\\
		    (y-2)(y-7) &= (2y+1)(y+1)\\
		    y^2 - 9y + 14 &= 2y^2 + 3y +1\\
		    -y^2 - 12y + 13&= 0\\
		    y^2+12y-13 &= 0\\
		    (y+13)(y-1)&=0\\
		    y = -13 &\text{ or } y  =1 
		    \end{align}\]
----
\(y=-13\)
----
\(y=-1\)
----
\(1 + \dfrac{t-2}{t-1} = \dfrac{5}{t^2-4t+3} + \dfrac{10}{3-t}\)
----
\(t \ne 1\)
----
\(t \ne 3\)
----
\[\begin{align}
 		    1 + \dfrac{t-2}{t-1} &= \dfrac{5}{t^2-4t+3} + \dfrac{10}{3-t} \\
		    1 + \dfrac{t-2}{t-1} &= \dfrac{5}{(t-3)(t-1)} - \dfrac{10}{t-3} \\
		    (t-3)(t-1) + (t-2)(t-3) &= 5 - 10(t-1) \\
		    t^2-4t+3 + t^2-5t+6 &= 5 - 10t + 10 \\
		    2t^2-9t+9 &= - 10t + 15\\
		    2t^2+t-6 &= 0\\
		    (2t-3)(t+2) &= 0\\
		    \text{Therefore } t = \frac{3}{2} &\text{ or } t  = -2
		    \end{align}\]
----
\(t=\frac{3}{2}\)
----
\(t=-2\)
----
\(\dfrac{4}{m+3} + \dfrac{4}{4-m^2} = \dfrac{5m-5}{m^2+m-6}\)
----
\(m \ne -3\)
----
\(m \ne \pm 2\)
----
\[\begin{align}
  		    \dfrac{4}{m+3} + \dfrac{4}{4-m^2} &= \dfrac{5m-5}{m^2+m-6}\\
		    \dfrac{4}{m+3} - \dfrac{4}{m^2-4} &= \dfrac{5(m-1)}{(m+3)(m-2)}\\
		    4(m^2-4)-4(m+3) &= 5(m-1)(m+2)\\
		    4m^2-16 - 4m - 12 &= 5(m^2+m-2)\\
		    4m^2- 4m - 28 &= 5m^2+5m-10\\
		    m^2+9m+18 &=0\\
		    (m+3)(m+6) &=0\\
		    m = -3  & \text{ or } m  = -6 \\
		    \text{But } m \ne -3, &\text{ therefore } m = -6
		    \end{align}\]
----
\(m=-6\)
----
\(5\sqrt{5t + 1} -4 = 5t + 1\)
----
\[\begin{align}
		   5\sqrt{5t + 1} -4 &= 5t + 1 \\
		   5\sqrt{5t + 1} &= 5t + 5 \\
		   5\sqrt{5t + 1} &= 5(t + 1) \\
		   \sqrt{5t + 1} &= t + 1 \\
		  \left(\sqrt{5t + 1} \right)^{2}  &= \left(t + 1 \right)^{2} \\
		  5t + 1  &= t^{2} + 2t + 1 \\
		  t^{2} - 3t  &= 0 \\
		  t(t - 3) & = 0 \\
		  t = 0 &\text{ or } t = 3
		  \end{align}\]
----
\(t= 0\)
----
\[\begin{align}
		    \text{LHS } &= 5\sqrt{5(0) + 1} -4 \\
		    &= 1 \\ 
		    \text{RHS } &= 5(0) + 1 \\ 
		    &= 1 \\
		    \therefore \text{LHS} & = \text{RHS}
		    \end{align}\]
----
\(t = 0\)
----
\(t = 3\)
----
\[\begin{align}
		    \text{LHS } &= 5\sqrt{5(3) + 1} -4  \\ 
		    &= 5\sqrt{16} - 4 \\ 
		    &= 16 \\ 
		    \text{RHS } &= 5(3) + 1  \\ 
		    &= 16 \\
		    \therefore \text{LHS} &= \text{RHS}
		    \end{align}\]
----
\(t = 3\)
----
\(t = 0\)
----
\(t = 3\)
----
\(x^2 - 4 = 0\)
----
\(x^2 - 8 = 0\)
----
\(x^2 -4x + 4 = 0\)
----
\(x^2 -4x - 4 = 0\)
----
\(x^2 - b^2\)
----
\((x-b)(x+b)\)
----
\(x^2-2x-1=0\)
----
\((x-1)^2\)
----
\((x-1)^2 = x^2-2x+1\)
----
\[\begin{align}
	x^2-2x-1 &= 0 \\ 
	(x^2-2x+1)-1-1 &= 0 \\ 
	(x^2-2x+1)-2 &= 0 \\ 
	(x-1)^2-2 &= 0 
	\end{align}\]
----
\(x\)
----
\[\begin{align}
	(x-1)^2-2 &= 0 \\
	(x-1)^2  &= 2 \\
	\sqrt{(x-1)^2} &= \pm \sqrt{2} \\
	x-1 &= \pm \sqrt{2} \\
	x &= 1 \pm \sqrt{2} \\
	\text{Therefore }x &= 1 + \sqrt{2} \text{ or }x = 1 - \sqrt{2}
	\end{align}\]
----
\(2^2 = 4\)
----
\((-2)^2 = 4\)
----
\(2 = \left(\sqrt{2}\right)^2\)
----
\[\begin{align}
	(x-1)^2-2 &= 0 \\ 
	(x-1)^2 - \left( \sqrt{2} \right)^2 &= 0 \\ 
	\left( (x-1) +  \sqrt{2} \right)\left( (x-1) -  \sqrt{2} \right) &= 0
	\end{align}\]
----
\[\begin{align}
	    (x-1) +  \sqrt{2} &= 0 \\ 
	    x &= 1 - \sqrt{2}
	    \end{align}\]
----
\[\begin{align}
	    (x-1) -  \sqrt{2} &= 0 \\ 
	    x &= 1 + \sqrt{2}
	    \end{align}\]
----
\(a{x}^{2}+bx+c=0\)
----
\({x}^{2}\)
----
\(a\)
----
\(x\)
----
\(x\)
----
\(x^2-10x-11=0\)
----
\(ax^2 + bx + c = 0\)
----
\(x^2\)
----
\[x^2-10x-11=0\]
----
\(x\)
----
\(x\)
----
\(x\)
----
\(x^2 - 10x + 25 - 25 - 11 = 0\)
----
\[\begin{align}
		  (x^2 - 10x + 25) - 25 - 11 &= 0 \\
		  (x-5)^2 - 36 &= 0
		\end{align}\]
----
\[\begin{align}
		  (x-5)^2 - 36 &= 0 \\
		  (x-5)^2  &= 36 \\
		  x-5 &= \pm\sqrt{36}
		\end{align}\]
----
\((6)^2 = 36\)
----
\((-6)^2 = 36\)
----
\[x - 5 = \pm 6\]
----
\(x\)
----
\[x = -1 \text{ or } x = 11\]
----
\[\begin{align}
		  (x-5)^2 - (6)^2 &= 0 \\
		  \left[\left(x-5\right) + 6\right] \left[\left(x-5\right) - 6\right] &= 0
		\end{align}\]
----
\(x\)
----
\[\begin{align}
		  (x+1)(x-11) &= 0 \\
		  \therefore x = -1 \text{ or } x &= 11
		\end{align}\]
----
\[x = -1 \text{ or } x = 11\]
----
\(2x^2 - 6x - 10 = 0\)
----
\(a{x}^{2}+bx+c=0\)
----
\(x^2\)
----
\({x}^{2}\)
----
\[x^2 - 3x - 5 = 0\]
----
\(x\)
----
\(x\)
----
\(\left( \dfrac{-3}{2} \right)^2 = \dfrac{9}{4}\)
----
\[\left( x^2 - 3x + \frac{9}{4}\right) - \frac{9}{4} - 5 = 0\]
----
\[\begin{align}
		\left( x - \frac{3}{2} \right)^2 - \frac{9}{4} - \frac{20}{4} &=  0 \\
		\left( x - \frac{3}{2} \right)^2 - \frac{29}{4} &=  0
		\end{align}\]
----
\[\begin{align}
		  \left( x - \frac{3}{2} \right)^2 - \frac{29}{4} &= 0 \\
		  \left( x - \frac{3}{2} \right)^2 &= \frac{29}{4} \\
		  x - \frac{3}{2} &= \pm \sqrt{\frac{29}{4}} 
		\end{align}\]
----
\(x\)
----
\[\begin{align}
		x - \frac{3}{2}  &= \pm \sqrt{\frac{29}{4}}  \\
		x &= \frac{3}{2} \pm \frac{\sqrt{29}}{2}  \\
		&= \frac{3 \pm \sqrt{29}}{2}
		\end{align}\]
----
\[\begin{align}
		\left( x - \frac{3}{2} \right)^2 - \frac{29}{4} &=  0 \\
		\left( x - \frac{3}{2} \right)^2 - \left( \sqrt{\frac{29}{4}} \right)^2 &=  0\\
		\left( x - \frac{3}{2}  - \sqrt{\frac{29}{4}} \right) \left( x - \frac{3}{2} + \sqrt{\frac{29}{4}} \right) &= 0
		\end{align}\]
----
\(x\)
----
\[\begin{align}
		\left( x - \frac{3}{2}  - \frac{\sqrt{29}}{2} \right) \left( x - \frac{3}{2} + \frac{\sqrt{29}}{2} \right) &= 0 \\
		\text{Therefore }  x = \frac{3}{2}  + \frac{\sqrt{29}}{2} &\text{ or } x = \frac{3}{2} - \frac{\sqrt{29}}{2} 
		\end{align}\]
----
\({x}^{2}+10x-2=0\)
----
\[\begin{align}
  x^2 + 10x - 2 &= 0\\
  x^2 +10x&= 2\\
  x^2 +10x +25 &= 2 + 25\\
  (x+5)^2 - 27 &= 0\\
  \left[(x+5) + \sqrt{27}\right]\left[(x+5) - \sqrt{27}\right] &= 0\\
  (x+5) = -\sqrt{27} &\text{ or } (x+5) = \sqrt{27}\\
  x = -5 - 3\sqrt{3} &\text{ or } x= -5 + 3\sqrt{3}
\end{align}\]
----
\(x=-5-3\sqrt{3}\)
----
\(x=-5 + 3\sqrt{3}\)
----
\({x}^{2}+4x+3=0\)
----
\[\begin{align}
  x^2+4x+3&=0\\
  x^2 +4x &= -3\\
  x^2 + 4x + 4 &= -3 + 4\\
  (x+2)^2 &= 1\\
  (x+2)&= \pm \sqrt {1}\\
  &=\pm 1\\
  x = -2 + 1 = -1 &\text{ or } x = -2-1=-3
\end{align}\]
----
\(x=-1\)
----
\(x=-3\)
----
\(p^2 - 5 = - 8p\)
----
\[\begin{align}
  p{^2} - 5 &= - 8p\\
  p{^2}+8p-5 &= 0\\
  p^2 + 8p &=5\\
  p^2 + 8p + 16 &= 5 + 16\\
  (p+4)^2 &= 21\\
  (p+4)&=\pm \sqrt{21}\\
  p &= -4 \pm \sqrt{21}
\end{align}\]
----
\(p=-4 \pm \sqrt{21}\)
----
\(2(6x + x^2) = -4\)
----
\[\begin{align}
	2(6x + x^2) &= -4 \\
  2x^2+12x+4 &= 0\\
  x^2 + 6x + 2 &= 0\\
  x^2 +6x &= -2\\
  x^2 + 6x + 9&= -2 +9\\
  (x+3)^2 &= 7\\
  x + 3 &= \pm \sqrt{7}\\
  x &= -3 \pm \sqrt{7}
\end{align}\]
----
\(x=-3 \pm \sqrt{7}\)
----
\({x}^{2}+5x+9=0\)
----
\[\begin{align}
  x^2+5x+9 &= 0\\
  x^2 + 5x &= -9\\
  x^2 + 5x + \frac{25}{4} &= -9 + \frac{25}{4}\\
  \left(x+\frac{5}{2}\right)^2 &= -\frac{11}{4}\\
  x + \frac{5}{2} &= \pm \sqrt{-\frac{11}{4}}\\
  \text{ No real solution }
\end{align}\]
----
\(t^2 + 30 = 2(10-8t)\)
----
\[\begin{align}
	t^2 + 30 &= 2(10 - 8t) \\
  t^2+16t+10&=0\\
  t^2 +16t &= -10\\
  t^2 + 16t+64 &= -10 +64\\
  (t+8)^2&=54\\
  t + 8 &= \pm \sqrt{54}\\
  t &= -8 \pm \sqrt{9 \times 6} \\
  \therefore t &= -8 \pm 3\sqrt{6}
\end{align}\]
----
\(t=-8 \pm 3\sqrt{6}\)
----
\(3{x}^{2}+6x-2=0\)
----
\[\begin{align}
  3x^2+6x-2&=0\\
  x^2 + 2x &= \frac{2}{3}\\
  x^2 + 2x + 1 &= \frac{2}{3} + 1\\
  (x+1)^2 &= \frac{5}{3}\\
  x +1 &= \pm \sqrt{\frac{5}{3}}\\
  x &= -1 \pm \sqrt{\frac{5}{3}}
\end{align}\]
----
\(x=-1 \pm \sqrt{\frac{5}{3}}\)
----
\({z}^{2}+8z-6=0\)
----
\[\begin{align}
  z^2 + 8z - 6 &= 0\\
  z^2 +8z &= 6\\
  z^2 + 8z+16 &= 6+16\\
  (z+4)^2 &= 22\\
  z+4 &= \pm \sqrt{22}\\
  z &= -4 \pm \sqrt{22}
\end{align}\]
----
\(z=-4 \pm \sqrt{22}\)
----
\(2z^2 = 11z\)
----
\[\begin{align}
  2z^2 &= 11z\\
  2z^2 - 11z &= 0\\
  z^2 - \frac{11}{2}z &= 0\\
  z^2 - \frac{11}{2}z + \frac{121}{16} &= \frac{121}{16}\\
  \left(z-\frac{11}{4}\right)^2 &= \frac{121}{16}\\
  z - \frac{11}{4} &= \pm \frac{11}{4}\\
  z = \frac{11}{4} + \frac{11}{4} = \frac{11}{2} &\text{ or } z = \frac{11}{4} - \frac{11}{4} = 0
\end{align}\]
----
\(z=\frac{11}{2}\)
----
\(z = 0\)
----
\(5+4z-{z}^{2}=0\)
----
\[\begin{align}
  5 + 4z - z^2 &= 0\\
  z^2 -4z &= 5\\
  z^2 - 4z +4 &= 5+4\\
  (z-2)^2 &= 9\\
  z-2 &= \pm \sqrt{9}\\
  z = 2+3 = 5 &\text{ or } z = 2-3 = -1
\end{align}\]
----
\(z= 5\)
----
\(z = -1\)
----
\(k\)
----
\(a\)
----
\(k^2 + 6k+ a = 0\)
----
\[\begin{align}
  k^2 + 6k+ a &= 0\\
  k^2 + 6k &= -a\\
  k^2 + 6k + 9 &= 9 - a\\
  (k + 3)^2 &= 9 - a\\
  k + 3 &= \pm \sqrt{9 - a}\\
  k & = -3 \pm \sqrt{9 - a} 
\end{align}\]
----
\(k = -3 \pm \sqrt{9 - a}\)
----
\(y\)
----
\(p\)
----
\(q\)
----
\(r\)
----
\(py^2 + qy + r = 0\)
----
\[\begin{align}
  py^2 + qy + r &= 0\\
  y^2 + \frac{q}{p}y + \frac{r}{p} &= 0\\
  y^2 + \frac{q}{p}y &= -\frac{r}{p}\\
  y^{2} + \frac{q}{p}y + \left(\frac{q}{2p}\right)^{2} &= \left(\frac{q}{2p}\right)^{2} - \frac{r}{p}\\
  \left(y + \frac{q}{2p}\right)^2 &= \frac{q^{2}}{4p^{2}} - \frac{r}{p} \\
  y + \frac{q}{2p} &= \pm \sqrt{\frac{q^{2} - 4pr}{4p^{2}}}\\
  y &=  -\frac{q}{2p} \pm \frac{\sqrt{q^{2} - 4pr}}{2p}\\
  y & = \frac{-q \pm \sqrt{q^{2} - 4pr}}{2p} 
\end{align}\]
----
\(y = \frac{-q \pm \sqrt{q^{2} - 4pr}}{2p}\)
----
\(ax^2 + bx + c = 0\)
----
\(a\)
----
\((a \ne 0)\)
----
\[x^2 + \frac{bx}{a} + \frac{c}{a} = 0\]
----
\(x\)
----
\(\left( \dfrac{b}{2a} \right)^2\)
----
\[\begin{align}
			x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} -\frac{b^2}{4a^2} + \frac{c}{a} &= 0 \\
			\left( x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} \right) -\frac{b^2}{4a^2} + \frac{c}{a} &= 0 \\
			\left( x + \frac{b}{2a} \right)^2 -\frac{b^2-4ac}{4a^2}  &= 0
			\end{align}\]
----
\[\begin{align}
		  	\left( x + \frac{b}{2a} \right)^2 &= \frac{b^2-4ac}{4a^2} \\
			\sqrt{\left( x + \frac{b}{2a} \right)^2} &= \pm \sqrt{\frac{b^2-4ac}{4a^2}} \\
			 x + \frac{b}{2a}  &= \pm \dfrac{\sqrt{b^2-4ac}}{2a} \\
			 x   &= - \dfrac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a} \\
			 x   &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} 
			\end{align}\]
----
\(ax^2+bx+c = 0\)
----
\[x = \dfrac{-b + \sqrt{b^2-4ac}}{2a} \text{  or  } x = \dfrac{-b - \sqrt{b^2-4ac}}{2a}\]
----
\({b}^{2}-4ac\)
----
\(x\)
----
\(2x^2+3x=7\)
----
\(a{x}^{2}+bx+c=0\)
----
\[2x^2+3x-7=0\]
----
\[a = 2; \qquad b = 3; \qquad c = -7\]
----
\(a\)
----
\(b\)
----
\(c\)
----
\[\begin{align}
  x &= \frac{-b±\sqrt{{b}^{2}-4ac}}{2a} \\
  &= \frac{-\left(3\right)±\sqrt{{\left(3\right)}^{2}-4\left(2\right)\left(-7\right)}}{2\left(2\right)} \\
  &= \frac{-3±\sqrt{65}}{4}
\end{align}\]
----
\(x=\dfrac{-3+\sqrt{65}}{4}\)
----
\(x=\dfrac{-3-\sqrt{65}}{4}\)
----
\(f(x)= x^2-5x+8\)
----
\(f(x)\)
----
\(x^2-5x+8 = 0\)
----
\[a = 1; \qquad b = -5; \qquad c = 8\]
----
\[\begin{align}
  x &= \frac{-b±\sqrt{{b}^{2}-4ac}}{2a} \\
  &= \frac{-\left(-5\right)±\sqrt{{\left(-5\right)}^{2}-4\left(1\right)\left(8\right)}}{2\left(1\right)} \\
  &= \frac{5±\sqrt{-7}}{2}
\end{align}\]
----
\(f(x)= x^2-5x+8\)
----
\(\sqrt{-7}\)
----
\(x\)
----
\(x\)
----
\(3{t}^{2}+t-4=0\)
----
\[\begin{align}
  3t^2 + t - 4 &= 0\\
  t &= \dfrac{-(1) \pm \sqrt{1^2 - 4(3)(-4)}}{2(3)}\\
  & = \dfrac{-1 \pm \sqrt{1+48}}{6}\\
  &=\dfrac{-1 \pm \sqrt{49}}{6}\\
  &=\dfrac{-1 \pm 7}{6}\\
  t = \dfrac{-1+7}{6}= \frac{6}{6}=1 &\text{ or } t = \dfrac{-1-7}{6}=\frac{-8}{6} = \frac{-4}{3}
\end{align}\]
----
\(t=1\)
----
\(t = \frac{-4}{3}\)
----
\({x}^{2}-5x-3=0\)
----
\[\begin{align}
  x^2 - 5x - 3 &= 0\\
  x &= \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-3)}}{2(1)}\\
  &= \dfrac{5 \pm \sqrt{25 + 12}}{2}\\
  &=\dfrac{5 \pm \sqrt{37}}{2}\\
  \text{therefore } x =\dfrac{5 + \sqrt{37}}{2} &\text{ or } x =\dfrac{5 - \sqrt{37}}{2}
\end{align}\]
----
\(x=\frac{5 + \sqrt{37}}{2}\)
----
\(t = \frac{5-\sqrt{37}}{2}\)
----
\(2{t}^{2}+6t+5=0\)
----
\[\begin{align}
  2t^2 + 6t + 5 &= 0\\
  t&=\dfrac{-6 \pm \sqrt{(6)^2 - 4(2)(5)}}{2(2)}\\
  &= \dfrac{-6 \pm \sqrt{36 - 40}}{4}\\
  &=\dfrac{-6 \pm \sqrt{-4}}{4}\\
  \text{No real solution}
  \end{align}\]
----
\(2p(2p+1)=2\)
----
\[\begin{align}
4p^2 + 2p - 2 &= 0\\
2p^2 + p - 1 &= 0\\
p &= \dfrac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)}\\
&= \dfrac{-1 \pm \sqrt{1+8}}{4}\\
&= \dfrac{-1 \pm \sqrt{9}}{4} \\
&= \dfrac{-1 \pm 3}{4} \\
  \text{therefore } p =\dfrac{-1 + 3}{4} = \frac{1}{2} &\text{ or } p =\dfrac{-1 - 3}{4} = -1
\end{align}\]
----
\(p = \frac{1}{2}\)
----
\(p = -1\)
----
\(-3{t}^{2}+5t-8=0\)
----
\[\begin{align}
-3t^2 + 5t -8 &= 0\\
t&= \dfrac{-5 \pm \sqrt{5^2 -4(-3)(-8)}}{2(-3)}\\
&= \dfrac{-5 \pm \sqrt{25-96}}{-6}\\
&=\dfrac{-5 \pm \sqrt{-71}}{-6}\\
\text{No real solution}
\end{align}\]
----
\(5{t}^{2}+3t-3=0\)
----
\[\begin{align}
5t^2 + 3t - 3 &= 0\\
t &= \dfrac{-3 \pm \sqrt{3^2 - 4(5)(-3)}}{2(5)}\\
&= \dfrac{-3 \pm \sqrt{9+60}}{10}\\
&=\dfrac{-3 \pm \sqrt{69}}{10}\\
  \text{therefore } t =\dfrac{-3 + \sqrt{69}}{10} &\text{ or } t =\dfrac{-3 - \sqrt{69}}{10}
\end{align}\]
----
\(t=\frac{-3+\sqrt{69}}{10}\)
----
\(t = \frac{-3 - \sqrt{69}}{10}\)
----
\({t}^{2}-4t+2=0\)
----
\[\begin{align}
t^2 - 4t + 2 &= 0\\
t &= \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)}\\
&= \dfrac{4 \pm \sqrt{16-8}}{2}\\
&=\dfrac{4 \pm \sqrt{8}}{2}\\
&= 2 \pm \sqrt{2}
\end{align}\]
----
\(t=2 \pm \sqrt{2}\)
----
\(9(k^2 - 1) = 7k\)
----
\[\begin{align}
9k^2 - 7k - 9 &= 0\\
k &= \dfrac{-(-7) \pm \sqrt{(-7)^2 - 4(9)(-9)}}{2(9)}\\
&= \dfrac{7 \pm \sqrt{49 + 324}}{18}\\
&=\dfrac{7 \pm \sqrt{373}}{18}\\
k = \dfrac{7 + \sqrt{373}}{18} &\text{ or } k =\dfrac{7 - \sqrt{373}}{18}
\end{align}\]
----
\(k=\frac{7+\sqrt{373}}{18}\)
----
\(k = \frac{7-\sqrt{373}}{18}\)
----
\(3f -2 = -2f^2\)
----
\[\begin{align}
2f^2 + 3f - 2 &= 0\\
f &= \dfrac{-3 \pm \sqrt{3^2-4(2)(-2)}}{2(2)}\\
&= \dfrac{-3 \pm \sqrt{9+16}}{4}\\
&= \dfrac{-3 \pm \sqrt{25}}{4}\\
&= \dfrac{-3 \pm 5}{4}\\
  \text{therefore } f =\dfrac{-3 + 5}{4} = \frac{1}{2}&\text{ or } f =\dfrac{-3 - 5}{4} = -2
\end{align}\]
----
\(f=\frac{1}{2}\)
----
\(f = -2\)
----
\({t}^{2}+t+1=0\)
----
\[\begin{align}
t^2 + t +1 &= 0\\
t&= \dfrac{-1 \pm \sqrt{1-4(1)(1)}}{2(1)}\\
&= \dfrac{-1 \pm \sqrt{-3}}{2}\\
\text{No real solution}
\end{align}\]
----
\(x\)
----
\(x^2 - 2x - \dfrac{3}{x^2 - 2x} = 2\)
----
\(x\)
----
\(x\)
----
\(x \ne 0\)
----
\(x \ne 2\)
----
\(x^2 - 2x\)
----
\(k = x^2 - 2x\)
----
\[k - \frac{3}{k} = 2\]
----
\(k\)
----
\(k\)
----
\(k \ne 0\)
----
\(k\)
----
\[\begin{align}
		  								k - \frac{3}{k}&= 2 \\
		  								k^2 - 3 &= 2k \\
		  								k^2 - 2k - 3 &= 0 \\
		  								(k + 1)(k - 3)&= 0 \\
											\text{Therefore } k = -1   &\text{ or } k = 3 
		   								\end{align}\]
----
\(k\)
----
\(k\)
----
\(x\)
----
\(k = -1\)
----
\[\begin{align}
		  								x^2 - 2x &= -1 \\
		  								x^2 - 2x + 1  &= 0 \\
		  								(x - 1)(x - 1) &= 0 \\
											\text{Therefore } x &= 1   
		   								\end{align}\]
----
\(k = 3\)
----
\[\begin{align}
		  								x^2 - 2x &= 3 \\
		  								x^2 - 2x - 3  &= 0 \\
		  								(x + 1)(x - 3)&= 0 \\
											\text{Therefore } x = -1   &\text{ or } x = 3  
		   								\end{align}\]
----
\(x\)
----
\(x = -1\)
----
\(x = 1\)
----
\(x = 3\)
----
\(-24 = 10(x^2 + 5x) + (x^2 + 5x)^2\)
----
\(x^2 + 5x\)
----
\(k = x^2 + 5x\)
----
\[-24 = 10k + k^{2}\]
----
\(k\)
----
\[\begin{align}
		  								k^{2} + 10k + 24 &= 0 \\
		  								(k + 6)(k + 4) &= 0 \\
											\text{Therefore } k = -6   &\text{ or } k = -4 
		   								\end{align}\]
----
\(k\)
----
\(x\)
----
\(k = -6\)
----
\[\begin{align}
		  								x^2 + 5x &= -6 \\
		  								x^2 + 5x + 6  &= 0 \\
		  								(x + 2)(x + 3) &= 0 \\
										\text{Therefore } x = -2 & \text{ or } x = -3    
		   								\end{align}\]
----
\(k = -4\)
----
\[\begin{align}
		  								x^2 + 5x &= -4 \\
		  								x^2 + 5x + 4  &= 0 \\
		  								(x + 1)(x + 4) &= 0 \\
										\text{Therefore } x = -1 & \text{ or } x = -4   
		   								\end{align}\]
----
\(x = -1\)
----
\(x = -4\)
----
\(x = -2\)
----
\(x = -3\)
----
\((x^2 - 2x)^2 - 8 = 7(x^2 - 2x)\)
----
\(x^2 - 2x\)
----
\(k = x^2 - 2x\)
----
\[k^{2} - 8 = 7k\]
----
\(k\)
----
\[\begin{align}
		  								k^{2} - 7k - 8 &= 0 \\
		  								(k + 1)(k - 8) &= 0 \\
											\text{Therefore } k = -1   &\text{ or } k = 8 
		   								\end{align}\]
----
\(k\)
----
\(x\)
----
\(k = -1\)
----
\[\begin{align}
		  								x^2 - 2x &= -1 \\
		  								x^2 - 2x + 1  &= 0 \\
		  								(x - 1)(x - 1) &= 0 \\
										\text{Therefore } x &= 1
		   								\end{align}\]
----
\(k = 8\)
----
\[\begin{align}
		  								x^2 - 2x &= 8 \\
		  								x^2 - 2x - 8  &= 0 \\
		  								(x + 2)(x - 4) &= 0 \\
										\text{Therefore } x = -2 & \text{ or } x = 4   
		   								\end{align}\]
----
\(x = 1\)
----
\(x = 4\)
----
\(x = -2\)
----
\(x^2 + 3x - \dfrac{56}{x(x + 3)} = 26\)
----
\(x\)
----
\(x \ne 0\)
----
\(x \ne -3\)
----
\(x^2 + 3x\)
----
\(x^2 + 3x\)
----
\(k = x^2 + 3x\)
----
\[k - \frac{56}{k} = 26\]
----
\(k\)
----
\(k \ne 0\)
----
\(k\)
----
\[\begin{align}
		  								k - \frac{56}{k}&= 26 \\
		  								k^{2} - 56 &= 26k \\
		  								k^{2} - 26k - 56 &= 0 \\
		  								(k - 28)(k + 2)&= 0 \\
											\text{Therefore } k = 28   &\text{ or } k = -2 
		   								\end{align}\]
----
\(k\)
----
\(k\)
----
\(x\)
----
\(k = 28\)
----
\[\begin{align}
		  								x^2 + 3x &= 28 \\
		  								x^2 + 3x - 28  &= 0 \\
		  								(x + 7)(x - 4) &= 0 \\
											\text{Therefore } x = -7 & \text{ or } x = 4   
		   								\end{align}\]
----
\(k = -2\)
----
\[\begin{align}
		  								x^2 + 3x &= -2 \\
		  								x^2 + 3x + 2  &= 0 \\
		  								(x + 2)(x + 1)&= 0 \\
											\text{Therefore } x = -1   &\text{ or } x = -2  
		   								\end{align}\]
----
\(x\)
----
\(x = -7\)
----
\(x = 4\)
----
\(x = -1\)
----
\(x = -2\)
----
\(x^2 - 18 + x + \dfrac{72}{x^2 + x} = 0\)
----
\(x\)
----
\(x \ne 0\)
----
\(x \ne -1\)
----
\(x^2 + x\)
----
\(k = x^2 + x\)
----
\[k - 18 + \frac{72}{k} = 0\]
----
\(k\)
----
\(k \ne 0\)
----
\(k\)
----
\[\begin{align}
		  								k - 18 + \frac{72}{k}&= 0\\
		  								k^{2} - 18k + 72 &= 0 \\
		  								(k - 12)(k - 6)&= 0 \\
											\text{Therefore } k = 12   &\text{ or } k = 6 
		   								\end{align}\]
----
\(k\)
----
\(k\)
----
\(x\)
----
\(k = 12\)
----
\[\begin{align}
		  								x^2 + x &= 12 \\
		  								x^2 + x - 12  &= 0 \\
		  								(x + 4)(x - 3) &= 0 \\
											\text{Therefore } x = -4 & \text{ or } x = 3   
		   								\end{align}\]
----
\(k = 6\)
----
\[\begin{align}
		  								x^2 + x &= 6 \\
		  								x^2 + x - 6  &= 0 \\
		  								(x + 3)(x - 2)&= 0 \\
											\text{Therefore } x = -3   &\text{ or } x = 2  
		   								\end{align}\]
----
\(x\)
----
\(x = -4\)
----
\(x = 3\)
----
\(x = -3\)
----
\(x = 2\)
----
\(x^2 - 4x + 10  - 7(4x - x^2) = -2\)
----
\[\begin{align}
		  								x^{2} - 4x + 10 - 28x + 7x^{2} & = -2 \\
		  								8x^{2} - 32x + 12 &= 0 \\
										2x^{2} - 8x + 3 & = 0 \\
										x &= \dfrac{-(-8) \pm \sqrt{(-8)^{2} - 4(2)(3)}}{2(2)}\\
										&= \dfrac{8 \pm \sqrt{64 - 24}}{4}\\
										&= \dfrac{8 \pm \sqrt{40}}{4}\\
							\text{therefore } x =\dfrac{8 + \sqrt{40}}{4} &\text{ or } x =\dfrac{8 - \sqrt{40}}{4}
		   								\end{align}\]
----
\(x = \frac{8 \pm \sqrt{40}}{4}\)
----
\(\dfrac{9}{x^2 + 2x -12} = x^2 + 2x -12\)
----
\(\frac{-2 \pm \sqrt{52}}{2}\)
----
\(x\)
----
\(x \ne \frac{-2 + \sqrt{52}}{2}\)
----
\(x \ne \frac{-2 - \sqrt{52}}{2}\)
----
\(x^2 + 2x - 12\)
----
\(k = x^2 + 2x - 12\)
----
\[\frac{9}{k} = k\]
----
\(k\)
----
\(k \ne 0\)
----
\(k\)
----
\[\begin{align}
		  								\frac{9}{k}&= k\\
		  								9 &= k^{2} \\
		  								k &= \pm 3 \\
											\text{Therefore } k = -3   &\text{ or } k = 3 
		   								\end{align}\]
----
\(k\)
----
\(k\)
----
\(x\)
----
\(k = 3\)
----
\[\begin{align}
		  								x^2 + 2x - 12 &= 3 \\
		  								x^2 + 2x - 15  &= 0 \\
		  								(x + 5)(x - 3) &= 0 \\
										\text{Therefore } x = -5 & \text{ or } x = 3   
		   								\end{align}\]
----
\(k = -3\)
----
\[\begin{align}
		  								x^2 + 2x - 12 &= -3 \\
		  								x^2 + 2x - 9  &= 0 \\
		  								x &= \dfrac{-2 \pm \sqrt{(2)^{2} - 4(1)(-9)}}{2(1)}\\
										&= \dfrac{-2 \pm \sqrt{4 + 36}}{2}\\
										&= \dfrac{-2 \pm \sqrt{40}}{2}\\
										&= \dfrac{-2 \pm 2\sqrt{10}}{2}\\
										&= -1 \pm \sqrt{10}\\
							\text{therefore } x = -1 + \sqrt{10} &\text{ or } x = -1 - \sqrt{10} 
		   								\end{align}\]
----
\(x\)
----
\(x = -5\)
----
\(x = 3\)
----
\(x = -1 + \sqrt{10}\)
----
\(x = -1 - \sqrt{10}\)
----
\[x = 13 \text{ or } x = -5\]
----
\[x - 13 = 0 \text{ or } x + 5 = 0\]
----
\[(x-13)(x+5) = 0\]
----
\[x^2 - 8x - 65 = 0\]
----
\[2x^2 - 16x - 130 = 0\]
----
\[-3x^2 + 24x + 195 = 0\]
----
\(-\frac{3}{2}\)
----
\[x = 4 \text{ or } x = -\frac{3}{2}\]
----
\[x - 4 = 0 \text{ or } x + \frac{3}{2} = 0\]
----
\[x - 4 = 0 \text{ or } 2x + 3 = 0\]
----
\[(2x+3)(x-4) = 0\]
----
\(2x^2 - 5x - 12 = 0\)
----
\[x = 3 \text{ or } x = -2\]
----
\[x - 3 = 0 \text{ or } x + 2 = 0\]
----
\[(x - 3)(x + 2) = 0\]
----
\[x^2 - x - 6 = 0\]
----
\(x^{2} - x - 6 = 0\)
----
\(x\)
----
\((-4;0)\)
----
\((4;0)\)
----
\(x\)
----
\[x = -4 \text{ or } x = 4\]
----
\[x + 4 = 0 \text{ or } x - 4 = 0\]
----
\[(x + 4)(x - 4) = 0\]
----
\[x^2 - 16 = 0\]
----
\(x^{2} - 16 = 0\)
----
\(ax^2 + bx + c = 0\)
----
\(a, b\)
----
\(c\)
----
\(-\frac{1}{2}\)
----
\[x = -\frac{1}{2} \text{ or } x = 3\]
----
\[x + \frac{1}{2} = 0 \text{ or } x - 3 = 0\]
----
\[(2x + 1)(x - 3) = 0\]
----
\[2x^2 - 5x - 3 = 0\]
----
\(2x^{2} - 5x - 3 = 0\)
----
\(k\)
----
\(kx^2 - 7x + 4 = 0\)
----
\(x=1\)
----
\(k\)
----
\[\begin{align*}
			  k(1)^{2} - 7(1) + 4 & = 0 \\
			  k & = 3 
			  \end{align*}\]
----
\(3x^2 - 7x + 4 = 0\)
----
\[\begin{align*}
			  3x^{2} - 7x + 4 & = 0 \\
			  (x - 1)(3x - 4) & = 0 \\
			  x = 1 & \text{ or } x = \frac{3}{4} 
			  \end{align*}\]
----
\(k=3\)
----
\(x=\frac{3}{4}\)
----
\(2x^2 - 3x = p\)
----
\(2\frac{1}{2}\)
----
\(p\)
----
\(p\)
----
\[\begin{align*}
			  2 \left( \frac{5}{2} \right)^{2} - 3 \left( \frac{5}{2} \right) - p & = 0 \\
			  \frac{25}{2} - \frac{15}{2} & = p \\
			  p & = 5 
			  \end{align*}\]
----
\(2x^2 - 3x - 5 = 0\)
----
\[\begin{align*}
			  2x^{2} - 3x - 5 & = 0 \\
			  (2x - 5)(x + 1) & = 0 \\
			  x = 2\frac{1}{2} & \text{ or } x = -1
			  \end{align*}\]
----
\(p=5\)
----
\(x=-1\)
----
\(24{y}^{2}+61y-8=0\)
----
\[\begin{align*}
				24y^{2} + 61y - 8 & = 0 \\
				(8y - 1)(3y + 8) & = 0 \\
				y = \frac{1}{8} & \text{ or } y = -\frac{8}{3} 
				\end{align*}\]
----
\(y = \frac{1}{8}\)
----
\(y = -\frac{8}{3}\)
----
\(8x^2 + 16x = 42\)
----
\[\begin{align*}
				8x^{2} + 16x - 42 & = 0 \\
				4x^{2} + 8x - 21 & = 0 \\
				(2x - 3)(2x + 7) & = 0 \\
				x = \frac{3}{2} & \text{ or } x = -\frac{7}{2} 
				\end{align*}\]
----
\(x = \frac{3}{2}\)
----
\(x = -\frac{7}{2}\)
----
\(9t^2 = 24t - 12\)
----
\[\begin{align*}
				9t^{2} - 24t + 12 & = 0 \\
				3t^{2} - 8t + 4 & = 0 \\
				(3t - 2)(t - 2) & = 0 \\
				t = \frac{2}{3} & \text{ or } t = 2
				\end{align*}\]
----
\(t = \frac{2}{3}\)
----
\(t = 2\)
----
\(-5{y}^{2}+0y+5=0\)
----
\[\begin{align*}
				-5y^{2} + 5 & = 0 \\
				-5(y^{2} - 1) & = 0 \\
				y^{2} -1 & = 0 \\
				(y-1)(y+1) & = 0 \\
				y = 1 & \text{ or } y = -1
				\end{align*}\]
----
\(y=1\)
----
\(y=-1\)
----
\(3m^2 + 12 = 15m\)
----
\[\begin{align*}
				3m^{2} + 12 - 15m & = 0 \\
				m^{2} - 5m + 4 & = 0 \\
				m^{2} - 5m & = -4 \\
				m^{2} - 5m + \frac{25}{4} & = -4 + \frac{25}{4} \\
				\left(m - \frac{5}{2}\right)^{2} & = \frac{9}{4} \\
				m - \frac{5}{2} & = \pm \sqrt{\frac{9}{4}} \\
				m & = \frac{5}{2} \pm \frac{3}{2} \\
				m = 1 & \text{ or } m = 4 
				\end{align*}\]
----
\(m = 1\)
----
\(m = 4\)
----
\(49{y}^{2}+0y-25=0\)
----
\[\begin{align*}
				49y^{2} - 25 & = 0 \\
				(7y - 5)(7y + 5) & = 0 \\
				y = \frac{5}{7} & \text{ or } y = -\frac{5}{7}
				\end{align*}\]
----
\(y = \pm \frac{5}{7}\)
----
\(72 = 66w - 12w^2\)
----
\[\begin{align*}
				12w^{2} - 66w + 72 & = 0 \\
				2w^{2} - 11w + 12 & = 0 \\
				(2w - 3)(w - 4) & = 0 \\
				w = \frac{3}{2} & \text{ or } w = 4
				\end{align*}\]
----
\(w = \frac{3}{2}\)
----
\(w = 4\)
----
\(-40{y}^{2}+58y-12=0\)
----
\[\begin{align*}
				-40y^{2} + 58y - 12 & = 0 \\
				20y^{2} - 29y + 6 & = 0 \\
				(5y - 6)(4y - 1) & = 0 \\
				y = \frac{6}{5} & \text{ or } y = \frac{1}{4}
				\end{align*}\]
----
\(y = \frac{6}{5}\)
----
\(y = \frac{1}{4}\)
----
\(37n + 72 - 24n^2 = 0\)
----
\[\begin{align*}
				24n^2 - 37n - 72 & = 0 \\
				(3n - 8)(8n + 9) & = 0 \\
				n = \frac{8}{3} & \text{ or } n = -\frac{9}{8}
				\end{align*}\]
----
\(n = \frac{8}{3}\)
----
\(n = -\frac{9}{8}\)
----
\(6{y}^{2}+7y-24=0\)
----
\[\begin{align*}
				6{y}^{2}+7y-24 & = 0 \\
				(3y + 8)(2y - 3) & = 0 \\
				y = -\frac{8}{3} & \text{ or } y = \frac{3}{2}
				\end{align*}\]
----
\(y = -\frac{8}{3}\)
----
\(y = \frac{3}{2}\)
----
\(3 = x(2x - 5)\)
----
\[\begin{align}
  3 &= 2x^2 - 5x\\
  2x^2 - 5x - 3 &= 0\\
  (2x+1)(x-3)&=0\\
  x =-\frac{1}{2}& \text{ or } x = 3
\end{align}\]
----
\(x= -\frac{1}{2}\)
----
\(x= 3\)
----
\(-18{y}^{2}-55y-25=0\)
----
\(y= -\frac{5}{2}\)
----
\(y= -\frac{5}{9}\)
----
\(-25{y}^{2}+25y-4=0\)
----
\[\begin{align}
  -25y^2 + 25y - 4 &= 0\\
  25y^2 - 25y+4 &= 0\\
  y^2 - y &= -\frac{4}{25}\\
  y^2 - y + \frac{1}{4} &= -\frac{4}{25} + \frac{1}{4}\\
  \left(y-\frac{1}{2}\right)^2 &= \frac{-16+25}{100}\\
  \left(y-\frac{1}{2}\right)^2 &= \frac{9}{100}\\
  y-\frac{1}{2} &= \pm \sqrt{\frac{9}{100}}\\
  y-\frac{1}{2} &= \pm \frac{3}{10}\\
  y = \frac{1}{2}+\frac{3}{10} = \frac{4}{5} &\text{ or } y = \frac{1}{2} - \frac{3}{10} = \frac{1}{5}
\end{align}\]
----
\(y= \frac{4}{5}\)
----
\(y= \frac{1}{5}\)
----
\(8(1 - 4g^2) + 24g = 0\)
----
\[\begin{align}
  -32g^2 + 24g + 8 &= 0\\
  32g^2 - 24g - 8 &= 0\\
  4g^2 - 3g - 1 &= 0\\
  (4g + 1)(g-1) &= 0\\
  g =-\frac{1}{4} &\text{ or } g=1
\end{align}\]
----
\(g= -\frac{1}{4}\)
----
\(g= 1\)
----
\(9{y}^{2}-13y-10=0\)
----
\[\begin{align}
  9y^2 - 13y - 10&= 0\\
  y &= \dfrac{-(-13) \pm \sqrt{(-13)^2 -4(9)(-10)}}{2(9)}\\
  &= \dfrac{-(-13) \pm \sqrt{169 + 360}}{18}\\
  &= \dfrac{13\pm \sqrt{529}}{18}\\
  &= \dfrac{13\pm23}{18}\\
  y = \dfrac{13+23}{18} = 2 &\text{ or } y = \dfrac{13-23}{18} = -\frac{5}{9}
\end{align}\]
----
\(y= 2\)
----
\(y= -\frac{5}{9}\)
----
\((7p - 3)(5p + 1) = 0\)
----
\[\begin{align}
  (7p-3)(5p+1)&=0\\
  p =\frac{3}{7} &\text{ or } p = -\frac{1}{5}
\end{align}\]
----
\(p= \frac{3}{7}\)
----
\(p= -\frac{1}{5}\)
----
\(-81{y}^{2}-99y-18=0\)
----
\[\begin{align}
  -81y^2 - 99y - 18 &= 0\\
  9y^2 + 11y +2 &=0\\
  (9y+2)(y+1) &= 0\\
  y = -\frac{2}{9} &\text{ or } y = -1
\end{align}\]
----
\(y= -\frac{2}{9}\)
----
\(y= -1\)
----
\(14{y}^{2}-81y+81=0\)
----
\(y= \frac{9}{2}\)
----
\(y= \frac{9}{7}\)
----
\(x^2 - 6x + 9 = 0\)
----
\(x^2 - 4x + 3 = 0\)
----
\(x^2 - 4x - 3 = 0\)
----
\(x^2 - 4x + 7 = 0\)
----
\(b^2 - 4ac\)
----
\(\Delta ={b}^{2}-4ac\)
----
\(Δ\)
----
\(f\left(x\right)=a{x}^{2}+bx+c\)
----
\(f\left(x\right)=0\)
----
\(x=\dfrac{-b±\sqrt{{b}^{2}-4ac}}{2a}=\dfrac{-b±\sqrt{\Delta }}{2a}\)
----
\(Δ < 0\)
----
\(Δ \geq 0\)
----
\(Δ = 0\)
----
\(Δ > 0\)
----
\(Δ\)
----
\(Δ\)
----
\(a>0\)
----
\(a<0\)
----
\(\Delta <0\)
----
\(\Delta=0\)
----
\(Δ > 0\)
----
\(Δ =\)
----
\(Δ =\)
----
\(x^2-2x-7=0\)
----
\(Δ\)
----
\(ax^2+bx+c=0\)
----
\[x^2-2x-7=0\]
----
\[a = 1; \qquad b = -2; \qquad c = -7\]
----
\[\begin{align}
		  Δ &= b^2-4ac \\
		   &= (-2)^2-4(1)(-7) \\
		   &= 4+28 \\
		   &= 32	
		   \end{align}\]
----
\(32 > 0\)
----
\(x^2 - 2x - 7 = 0\)
----
\(Δ  > 0\)
----
\(k\)
----
\(6x^2 + 6 = 4kx\)
----
\(k\)
----
\(\Delta =0\)
----
\(ax^2+bx+c=0\)
----
\[6x^2-4kx+6=0\]
----
\[a = 6; \qquad b = -4k; \qquad c = 6\]
----
\[\begin{align}
		  Δ &= b^2-4ac \\
		   &= (-4k)^2-4(6)(6) \\
		   &= 16k^2-144 
		   \end{align}\]
----
\(\Delta =0\)
----
\[\begin{align}
		  Δ &= 0 \\
		   16k^2 - 144 &= 0 \\
		   16(k^2-9)&= 0 \\ 
		   (k-3)(k+3)&= 0 
		   \end{align}\]
----
\(k = 3\)
----
\(k = -3\)
----
\(k = 3\)
----
\[\begin{align}
		  6x^2-4(3)x + 6 &= 0 \\
		  6x^2-12x + 6 &= 0 \\
		   x^2-2x + 1 &= 0 \\
		   (x-1)(x-1)&= 0 \\ 
		   (x-1)^2&= 0 \\ 
		  \text{Therefore } x &= 1
		   \end{align}\]
----
\(k = 3\)
----
\(x = 1\)
----
\(k = - 3\)
----
\[\begin{align}
		  6x^2-4(-3)x + 6 &= 0 \\
		  6x^2+12x + 6 &= 0 \\
		   x^2+2x + 1 &= 0 \\
		   (x+1)(x+1)&= 0 \\ 
		   (x+1)^2&= 0 \\ 
		  \text{Therefore } x &= -1
		   \end{align}\]
----
\(k = -3\)
----
\(x = -1\)
----
\(k = 3\)
----
\(k = -3\)
----
\(\)
----
\((x+h)(x+k)=4d^2\)
----
\(h, k\)
----
\(d\)
----
\(Δ\)
----
\(Δ \ge 0\)
----
\(h\)
----
\(k\)
----
\(d\)
----
\(ax^2+bx+c=0\)
----
\[\begin{align}
		  (x+h)(x+k) &= 4d^2 \\
		  x^2 + hx +kx +hk -4d^2 &= 0 \\
		  x^2 + (h+k)x + (hk -4d^2) &= 0
		  \end{align}\]
----
\[a = 1; \qquad b = h+k; \qquad c = hk-4d^2\]
----
\[\begin{align}
		  \Delta &= b^2-4ac \\
		  &= (h+k)^2-4(1)(hk-4d^2) \\
		  &= h^2+2hk+k^2-4hk+16d^2 \\
		  &= h^2-2hk+k^2+16d^2 \\
		  &= (h - k)^2+(4d)^2
		\end{align}\]
----
\(\Delta \geq 0\)
----
\[\begin{align}
		    \text{We know that } (4d)^2 &\geq 0 \\
		    \text{and }     (h - k)^2 &\geq 0 \\
		    \text{so then }   (h - k)^2+ (4d)^2 &\geq 0 \\
		    \text{therefore }  \Delta &\geq 0
		 	\end{align}\]
----
\(\Delta \geq 0\)
----
\(h, k\)
----
\(d\)
----
\(x^2 + 3x = -2\)
----
\(ax^2+bx+c=0\)
----
\[x^2 + 3x + 2 = 0\]
----
\[a = 1; \qquad b = 3; \qquad c = 2\]
----
\[\begin{align}
		  Δ &= b^2-4ac \\
		   &= (3)^2 - 4(1)(2) \\
		   &= 9 - 8 \\
		   &= 1
		   \end{align}\]
----
\(1 > 0\)
----
\(Δ  > 0\)
----
\(x^2 + 9 = 6x\)
----
\(ax^2+bx+c=0\)
----
\[x^2 - 6x + 9 = 0\]
----
\[a = 1; \qquad b = -6; \qquad c = 9\]
----
\[\begin{align}
		  Δ &= b^2-4ac \\
		   &= (-6)^2 - 4(1)(9) \\
		   &= 36 - 36 \\
		   &= 0
		   \end{align}\]
----
\(Δ = 0\)
----
\(6y^2 - 6y - 1 = 0\)
----
\(ax^2+bx+c=0\)
----
\[6y^2 - 6y - 1 = 0\]
----
\[a = 6; \qquad b = -6; \qquad c = -1\]
----
\[\begin{align}
		  Δ &= b^2-4ac \\
		   &= (-6)^2 - 4(6)(-1) \\
		   &= 36 + 36 \\
		   &= 72
		   \end{align}\]
----
\(72 > 0\)
----
\(Δ > 0\)
----
\(4t^2 - 19t - 5 = 0\)
----
\(ax^2+bx+c=0\)
----
\[4t^2 - 19t - 5 = 0\]
----
\[a = 4; \qquad b = -19; \qquad c = -5\]
----
\[\begin{align}
		  Δ &= b^2-4ac \\
		   &= (-19)^2 - 4(4)(-5) \\
		   &= 361 + 80 \\
		   &= 441
		   \end{align}\]
----
\(441 > 0\)
----
\(Δ > 0\)
----
\(z^2 = 3\)
----
\(ax^2+bx+c=0\)
----
\[z^2 - 3 = 0\]
----
\[a = 1; \qquad b = 0; \qquad c = -3\]
----
\[\begin{align}
		  Δ &= b^2-4ac \\
		   &= (0)^2 - 4(1)(-3) \\
		   &= 0 + 12 \\
		   &= 12
		   \end{align}\]
----
\(12 > 0\)
----
\(Δ > 0\)
----
\(0 = p^2 + 5p + 8\)
----
\(ax^2+bx+c=0\)
----
\[p^2 + 5p + 8 = 0\]
----
\[a = 1; \qquad b = 5; \qquad c = 8\]
----
\[\begin{align}
		  Δ &= b^2-4ac \\
		   &= (5)^2 - 4(1)(8) \\
		   &= 25 - 32 \\
		   &= -7
		   \end{align}\]
----
\(-7 < 0\)
----
\(Δ < 0\)
----
\(x^2 = 36\)
----
\(ax^2+bx+c=0\)
----
\[x^2 - 36 = 0\]
----
\[a = 1; \qquad b = 0; \qquad c = -36\]
----
\[\begin{align}
		  Δ &= b^2-4ac \\
		   &= (0)^2 - 4(1)(-36) \\
		   &= 0 + 144 \\
		   &= 144
		   \end{align}\]
----
\(144 > 0\)
----
\(Δ > 0\)
----
\(4m + m^2 = 1\)
----
\(ax^2+bx+c=0\)
----
\[m^2 + 4m - 1 = 0\]
----
\[a = 1; \qquad b = 4; \qquad c = -1\]
----
\[\begin{align}
		  Δ &= b^2-4ac \\
		   &= (4)^2 - 4(1)(-1) \\
		   &= 16 + 4 \\
		   &= 20
		   \end{align}\]
----
\(20 > 0\)
----
\(Δ  > 0\)
----
\(11 - 3x + x^2 = 0\)
----
\(ax^2+bx+c=0\)
----
\[x^2 - 3x + 11 = 0\]
----
\[a = 1; \qquad b = -3; \qquad c = 11\]
----
\[\begin{align}
		  Δ &= b^2-4ac \\
		   &= (-3)^2 - 4(1)(11) \\
		   &= 9 - 44 \\
		   &= -35
		   \end{align}\]
----
\(-35 < 0\)
----
\(Δ  < 0\)
----
\(y^2 + \dfrac{1}{4} = y\)
----
\(ax^2+bx+c=0\)
----
\[4y^2 - 4y + 1 = 0\]
----
\[a = 4; \qquad b = -4; \qquad c = 1\]
----
\[\begin{align}
		  Δ &= b^2-4ac \\
		   &= (-4)^2 - 4(4)(1) \\
		   &= 16 - 16 \\
		   &= 0
		   \end{align}\]
----
\(Δ = 0\)
----
\({x}^{2}+bx-2+k\left({x}^{2}+3x+2\right)=0\)
----
\(\left(k\ne -1\right)\)
----
\(\Delta ={k}^{2}+6bk+{b}^{2}+8\)
----
\(ax^2+bx+c=0\)
----
\[(k + 1)x^2 + (b + 3k)x - 2 + 2k = 0\]
----
\[a = k + 1; \qquad b = b + 3k; \qquad c = -2 + 2k\]
----
\[\begin{align}
		  Δ &= b^2-4ac \\
		   &= (b + 3k)^2 - 4(k + 1)(2k - 2) \\
		   &= b^{2} + 6bk + 9k^{2} -4(2k^{2} - 2) \\
		   &= b^{2} + 6bk + 9k^{2} -8k^{2} + 8 \\
		   &= k^{2} + 6bk + b^{2} + 8
		   \end{align}\]
----
\(b=0\)
----
\(b=0\)
----
\[Δ = k^{2} + 8\]
----
\(k\)
----
\(k\)
----
\(k = 0\)
----
\(Δ = 8\)
----
\(k = -1\)
----
\(Δ = 9\)
----
\(k = 1\)
----
\(Δ = 9\)
----
\(k\)
----
\(b=2\)
----
\(k\)
----
\(b=2\)
----
\[\begin{align}
		  Δ & = k^{2} + 6(2)k + (2)^{2} + 8 \\
		  & = k^{2} + 12k + 12 
		  \end{align}\]
----
\(k\)
----
\[\begin{align}
		  0 & = k^{2} + 12k + 12 \\
		  k & = \dfrac{-12 \pm \sqrt{(12)^{2} - 4(1)(12)}}{2(1)} \\
		  & = \dfrac{-12 \pm \sqrt{144 - 48}}{2} \\
		  & = \dfrac{-12 \pm \sqrt{96}}{2} \\
		  & = \dfrac{-12 \pm 4\sqrt{6}}{2} \\
		  k = -6 + 2\sqrt{6}  & \text{ or } k =  -6 - 2\sqrt{6} 
		  \end{align}\]
----
\(k = -6 \pm 2\sqrt{6}\)
----
\({k}^{2}{x}^{2}+2=kx-{x}^{2}\)
----
\(k\)
----
\[\begin{align}
      k^2 x^2 + x^2 - kx +2 &= 0\\
      a &= (k^2+1)\\
      b&= -k\\
      c&=2\\
\Delta &= b^2-4ac\\
      &= (-k)^2 - 4(k^2+1)(2)\\
      &=k^2-8k^2-8\\
      &=-7k^2 - 8\\
      &=-(7k^2 + 8)\\
      \Delta & < 0
      \end{align}\]
----
\({x}^{2}+12x=3k{x}^{2}+2\)
----
\(k\)
----
\(k \in \mathbb{Z}\)
----
\[\begin{align}
x^2 +12x &= 3kx^2 +2\\
3kx^2 -x^2 - 12x+2&=0\\
x^2(3k-1)-12x+2&=0\\
a&=3k-1\\
b&=-12\\
c&=2\\
\text{Given } \Delta &\geq 0\\
\therefore b^2-4ac &\geq 0\\
\therefore (-12)^2 - 4(3k-1)(2) &\geq 0\\
144 -24k + 8 &\geq 0 \\
152 - 24k &\geq 0 \\
152 &\geq 24k \\
 \frac{19}{3} &\geq k \\
\therefore k &\leq \frac{19}{3} \\
\therefore k &\leq 6\frac{1}{3}
\end{align}\]
----
\(k\)
----
\(k\)
----
\(k=6\)
----
\(k\)
----
\(\Delta\)
----
\[\begin{align}
\Delta &= b^2-4ac \\
&=152-24k\\
\text{if } k&=\frac{1}{3}\\
\Delta &= 152 - 24\left(\frac{1}{3}\right)\\
&=152-8\\
&=144\\
&=12^2
\end{align}\]
----
\(k=\frac{1}{3}\)
----
\[k = \frac{x^2-4}{2x-5}\]
----
\(x \ne \frac{5}{2}\)
----
\(k\)
----
\[\begin{align}
				  k(2x - 5) & = x^2-4 \\
				  2kx - 5k & = x^2-4 \\
				  0 & = x^{2} - 4 - 2kx + 5k  \\
				  0 & = x^{2} - 2kx + 5k - 4  
				  \end{align}\]
----
\(a = 1; \qquad b = -2k; \qquad c = 5k - 4\)
----
\[\begin{align}
		  Δ &= b^2-4ac \\
		   &= (-2k)^2 - 4(1)(5k - 4) \\
		   &= 4k^{2} - 20k + 16
		   \end{align}\]
----
\(k\)
----
\(k\)
----
\[\begin{align}
		  0 &= 4k^{2} - 20k + 16 \\
		   &= k^{2} - 5k + 4 \\
		   &= (k - 4)(k - 1) \\
		   k = 4 & \text{ or } k = 1
		   \end{align}\]
----
\(k=4\)
----
\(k=1\)
----
\(k\)
----
\[\Delta = 4k^{2} - 20k + 16\]
----
\(k\)
----
\[\begin{align}
		   \Delta & = 4k^{2} - 20k + 16 \\
		   1 & = 4k^{2} - 20k + 16 \\
		   0 & = 4k^{2} - 20k + 15
		   \end{align}\]
----
\(k\)
----
\[\begin{align}
		   4 & = 4k^{2} - 20k + 16 \\
		   0 & = 4k^{2} - 20k + 12 \\
		   & = k^{2} - 5k + 3 
		   \end{align}\]
----
\(k\)
----
\[\begin{align}
		   9 & = 4k^{2} - 20k + 16 \\
		   0 & = 4k^{2} - 20k + 5
		   \end{align}\]
----
\(k\)
----
\[\begin{align}
		   16 & = 4k^{2} - 20k + 16 \\
		   0 & = 4k^{2} - 20k \\
		   & = k^{2} - 5k \\
		   k = 0 & \text{ or } k = 5 
		   \end{align}\]
----
\(k=0\)
----
\(k=5\)
----
\({x}^{2}-\left(a+b\right)x+ab-{p}^{2}=0\)
----
\(a\)
----
\(b\)
----
\(p\)
----
\(\Delta \geq 0\)
----
\[\begin{align}
		      \Delta
		      &=(-a-b)^2 - 4(ab-p^2)\\
		      &=a^2 + 2ab + b^2 - 4ab + 4p^2\\
		      &=a^2 - 2ab + b^2 + 4p^2\\
		      &=(a-b)^2 + 4p^2
		      \end{align}\]
----
\(\Delta \geq 0\)
----
\(a\)
----
\(b\)
----
\(p\)
----
\(a\)
----
\(b\)
----
\(p\)
----
\(\Delta = 0\)
----
\(a=b\)
----
\(p=0\)
----
\(b\)
----
\(c\)
----
\(b;\phantom{\rule{0.222222em}{0ex}}c\)
----
\({x}^{2}+bx+c=0\)
----
\(a\)
----
\(b\)
----
\(c\)
----
\(\Delta \geq 0\)
----
\((b;c)\)
----
\((1;1),~(1;2),~(1;3),~(2;1),~(2;2),~(2;3),~(3;1),~(3;2),~(3;3)\)
----
\(\Delta\)
----
\((\Delta < 0),~(\Delta < 0),~(\Delta < 0),~(\Delta = 0),~(\Delta < 0),~(\Delta < 0),~(\Delta >0),~(\Delta > 0),~(\Delta < 0)\)
----
\(\Delta ≥ 0\)
----
\((b;c) = (2;1),~(3;1),~(3;2)\)
----
\[\begin{align}
  ax^2 + bx + c > 0 \\
  ax^2 + bx + c \ge 0 \\
  ax^2 + bx + c < 0 \\
  ax^2 + bx + c \le 0
\end{align}\]
----
\(x\)
----
\(x\)
----
\({x}^{2}-5x+6\ge 0\)
----
\[(x-3)(x-2) \geq 0\]
----
\(x\)
----
\(x=3\)
----
\(x=2\)
----
\(x=2\)
----
\(x=3\)
----
\(x-3\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(+\)
----
\(x-2\)
----
\(-\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(f(x)=(x-3)(x-2)\)
----
\(+\)
----
\(-\)
----
\(+\)
----
\(f(x)\)
----
\(x\le 2\)
----
\(x\ge 3\)
----
\(x\)
----
\(x^2 - 5x + 6\)
----
\(a>0\)
----
\((x-3)(x-2)\)
----
\(x\)
----
\((2;0)\)
----
\((3;0)\)
----
\(x\)
----
\(x \le 2\)
----
\(x \ge 3\)
----
\[x^2 - 5x + 6 \geq 0 \text{ for } x \leq 2 \text{ or } x \geq 3\]
----
\(x\)
----
\(4{x}^{2}-4x+1\le 0\)
----
\[\begin{align}
		  (2x-1)(2x-1) & \leq 0 \\
		  (2x-1)^2 & \leq 0
		   \end{align}\]
----
\(x\)
----
\(x=\frac{1}{2}\)
----
\(a^2 > 0\)
----
\(a, a \ne 0\)
----
\((2x-1)^2\)
----
\(4x^2 - 4x + 1\)
----
\(a > 0\)
----
\((2x-1)(2x-1)\)
----
\(x\)
----
\(\left(\frac{1}{2};0\right)\)
----
\(x\)
----
\[4x^2 - 4x + 1 \leq 0 \text{ for }  x = \frac{1}{2}\]
----
\(x\)
----
\(-{x}^{2}-3x+5>0\)
----
\(x^2\)
----
\[-x^2-3x+5 > 0\]
----
\[x^2+3x-5 < 0\]
----
\(x\)
----
\(x\)
----
\(-{x}^{2}-3x+5>0\)
----
\(-x^2-3x+5=0\)
----
\[\begin{align}
  -x^2 - 3x + 5 &= 0 \\
  x^2 + 3x - 5 &= 0
\end{align}\]
----
\(x\)
----
\(-\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(+\)
----
\(+\)
----
\(-\)
----
\(+\)
----
\(x^2+3x-5\)
----
\(a > 0\)
----
\(x\)
----
\(x\)
----
\(x\)
----
\(\dfrac{2}{x+3} = \dfrac{1}{x-3}\)
----
\(x \ne \pm 3\)
----
\(\dfrac{2}{x+3} \le \dfrac{1}{x-3}\)
----
\(x \ne \pm 3\)
----
\((x+3)(x-3)\)
----
\[\begin{align}
		  \frac{2}{x+3} \times (x+3)(x-3) &= \frac{1}{x-3} \times (x+3)(x-3)\\
		  2(x-3) &= x+3\\
		  2x-6 &= x+3\\
		  x &= 9
		  \end{align}\]
----
\(\dfrac{1}{x-3}\)
----
\[\frac{2}{x+3} - \frac{1}{x-3} \le 0\]
----
\[\begin{align}
		  \frac{2(x-3) - (x+3)}{(x+3)(x-3)} &\le 0 \\
		  \frac{x-9}{(x+3)(x-3)} &\le 0
		\end{align}\]
----
\(x\)
----
\(x=-3\)
----
\(x=3\)
----
\(x=9\)
----
\(x=-3\)
----
\(x=3\)
----
\(x=9\)
----
\(x+3\)
----
\(-\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(x-3\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(x-9\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(+\)
----
\(f(x)=\dfrac{x-9}{(x+3)(x-3)}\)
----
\(-\)
----
\(+\)
----
\(-\)
----
\(+\)
----
\(x < -3\)
----
\(3 < x \leq 9\)
----
\(x = -3\)
----
\(x = 3\)
----
\[x < -3 \quad\text{or}\quad 3 < x \le 9\]
----
\({x}^{2}-x<12\)
----
\[\begin{align}
				x^{2} - x - 12 & < 0 \\
				(x - 4)(x + 3) & < 0
				\end{align}\]
----
\(x=-3\)
----
\(x=4\)
----
\(x - 4\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(+\)
----
\(x + 3\)
----
\(-\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(f(x)= (x - 4)(x + 3)\)
----
\(+\)
----
\(-\)
----
\(+\)
----
\(f(x)\)
----
\(-3 < x < 4\)
----
\(-3 < x < 4\)
----
\(3{x}^{2}>-x+4\)
----
\[\begin{align}
				3x^{2} + x - 4 & > 0 \\
				(3x + 4)(x - 1) & > 0
				\end{align}\]
----
\(x=-\frac{4}{3}\)
----
\(x=1\)
----
\(x - 1\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(+\)
----
\(3x + 4\)
----
\(-\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(f(x)= (3x + 4)(x - 1)\)
----
\(+\)
----
\(-\)
----
\(+\)
----
\(f(x)\)
----
\(x < -\frac{4}{3}\)
----
\(x > 1\)
----
\({y}^{2}<-y-2\)
----
\[y^{2} + y + 2 < 0\]
----
\(x\)
----
\(x\)
----
\(y\)
----
\((3-t)(1+t) > 0\)
----
\[(3 - t)(1 + t) > 0\]
----
\(t = -1\)
----
\(t = 3\)
----
\(3 - t\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(-\)
----
\(1 + t\)
----
\(-\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(f(x)= (3 - t)(1 + t)\)
----
\(-\)
----
\(+\)
----
\(-\)
----
\(f(x)\)
----
\(-1 < t < 3\)
----
\({s}^{2}-4s>-6\)
----
\[s^{2} - 4s + 6 > 0 \\\]
----
\[\begin{align}
				s & = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(6)}}{2(1)} \\
				 & = \frac{4 \pm \sqrt{16- 24}}{2 } \\
				 & = \frac{4 \pm \sqrt{-8}}{2} 
				\end{align}\]
----
\(x\)
----
\(x\)
----
\(s\)
----
\(s\)
----
\(0 \ge 7{x}^{2}-x+8\)
----
\[\begin{align}
				x & = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(7)(8)}}{2(7)} \\
				 & = \frac{1 \pm \sqrt{1- 224}}{14} \\
				 & = \frac{1 \pm \sqrt{-223}}{14} 
				\end{align}\]
----
\(x\)
----
\(x\)
----
\(x\)
----
\(x\)
----
\(x \geq -4x^2\)
----
\[\begin{align}
		  x & \geq -4x^2 \\
		  4x^2 + x & \geq 0 \\
		  x(4x + 1) & \geq 0
		  \end{align}\]
----
\(x = -\frac{1}{4}\)
----
\(x = 0\)
----
\(x\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(+\)
----
\(4x + 1\)
----
\(-\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(f(x)= x(4x + 1)\)
----
\(+\)
----
\(-\)
----
\(+\)
----
\(f(x)\)
----
\(x \leq -\frac{1}{4}\)
----
\(x \geq 0\)
----
\(2{x}^{2}+x+6\le 0\)
----
\[2x^{2} + x + 6  \le 0\]
----
\(x\)
----
\(x\)
----
\(\dfrac{x}{x-3}<2\)
----
\(x\ne 3\)
----
\[\begin{align}
		  \dfrac{x}{x-3} & < 2 \\
		  \dfrac{x}{x-3} - 2 & < 0 \\
		  \dfrac{x - 2(x-3)}{x-3}& < 0 \\
		  \dfrac{x - 2x + 6}{x-3}& < 0 \\
		  \dfrac{-x + 6}{x-3}& < 0 \\
		  \dfrac{-(x - 6)}{x-3}& < 0 \\
		  \dfrac{x - 6}{x-3}& > 0 \\
		  x & = 6 
		  \end{align}\]
----
\(x = 3\)
----
\(x = 6\)
----
\(x - 3\)
----
\(-\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(x - 6\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(+\)
----
\(f(x)= x - 6\)
----
\(+\)
----
\(-\)
----
\(+\)
----
\(f(x) > 0\)
----
\(x < 3 \text{ or } x > 6\)
----
\(x \ne 3\)
----
\(\dfrac{x^2 + 4}{x - 7} \geq 0\)
----
\(x \ne 7\)
----
\[\dfrac{(x + 2)(x - 2)}{x - 7} \geq 0\]
----
\(x = -2\)
----
\(x = 2\)
----
\(x = 7\)
----
\(x + 2\)
----
\(-\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(x - 2\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(x - 7\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(+\)
----
\(f(x)= \dfrac{(x + 2)(x - 2)}{x - 7}\)
----
\(-\)
----
\(+\)
----
\(-\)
----
\(+\)
----
\(f(x)\)
----
\(-2 \leq x \leq 2\)
----
\(x > 7\)
----
\(x \ne 7\)
----
\(\dfrac{x + 2}{x} - 1 \geq 0\)
----
\(x \ne 0\)
----
\[\begin{align}
		\frac{x + 2}{x} - 1 & \ge 0 \\
		\frac{x + 2 - x}{x}& \ge 0 \\
		\frac{2}{x} & \ge 0 \\
		\text{Therefore  } x & > 0
		\end{align}\]
----
\(x > 0\)
----
\(x \ne 0\)
----
\(x\)
----
\(2x^2-18 > 0\)
----
\(x < -3\)
----
\(x > 3\)
----
\(5 - x^2 \leq 0\)
----
\[\begin{align}
		5 - x^2  & \leq 0 \\
		 - (x^2  - 5) & \leq 0 \\
		 x^2  - 5 & \geq 0 \\
		\text{Therefore  } x & > 0
		\end{align}\]
----
\(-\sqrt{5} \leq x \leq \sqrt{5}\)
----
\(x^2 < 0\)
----
\(x\)
----
\(0 \ge 6x^2\)
----
\(x\)
----
\(x\)
----
\(x\)
----
\(n\)
----
\(n\)
----
\[\begin{align}
			  y - 2x &= -4\\
			  x^2 + y &= 4 
		  \end{align}\]
----
\(x\)
----
\(y\)
----
\[\begin{alignrll}
		    y-2x &= -4 & \ldots (1) \\
		    x^2 + y &= 4 & \ldots (2) 
		  \end{alignrll}\]
----
\(y\)
----
\[y = 2x - 4\]
----
\[\begin{align}
		  x^2 + \left(2x-4\right) &= 4 \\
		  x^2 + 2x - 8 &= 0
		\end{align}\]
----
\[\begin{align}
		  \left(x+4\right)\left(x-2\right) &= 0 \\
		  \therefore x = -4 &\text{ or } x = 2
		\end{align}\]
----
\(x\)
----
\(y\)
----
\(x = -4\)
----
\[\begin{aligned}
		  y &=  2(-4)-4 \\
		  &= -12
		   \end{aligned}\]
----
\(x = 2\)
----
\[\begin{aligned}
		  y &=  2(2) - 4 \\
		  &= 0
		   \end{aligned}\]
----
\(x = -4 \text{ and } y = -12\)
----
\(x = 2 \text{ and } y = 0\)
----
\(x\)
----
\(y\)
----
\[\begin{alignrll}
	        y &= x^2 - 6x & \ldots (1) \\
	        y + \frac{1}{2}x - 3 &= 0 & \ldots (2) 
	      \end{alignrll}\]
----
\(y\)
----
\[\begin{align}
		 					y + \frac{1}{2}x - 3 &= 0 \\
								y  &= -\frac{1}{2}x + 3
		   				\end{align}\]
----
\(x\)
----
\[\begin{align}
		 					x^2 - 6x &= -\frac{1}{2}x + 3 \\
		 					x^2 - 6x  + \frac{1}{2}x - 3 &= 0\\
		 					2x^2 - 12x  + x - 6 &= 0\\
		 					2x^2 -11x - 6 &= 0\\
		 					(2x +1)(x-6) &= 0\\
		 					\text{Therefore } x = -\frac{1}{2}&\text{ or } x = 6
		   				\end{align}\]
----
\(x\)
----
\(y\)
----
\(x = -\dfrac{1}{2}\)
----
\[\begin{align}
		    y &= -\frac{1}{2} \left( -\frac{1}{2} \right) + 3  \\
		    \therefore y &= 3\frac{1}{4}
		    \end{align}\]
----
\(\left(-\dfrac{1}{2};3\dfrac{1}{4}\right)\)
----
\(x = 6\)
----
\[\begin{align}
		    y &= -\frac{1}{2}(6) + 3 \\
		    &= -3 + 3 \\
		    \therefore y &= 0
		    \end{align}\]
----
\((6;0)\)
----
\(x = -\dfrac{1}{2} \text{ and } y = 3\dfrac{1}{4}\)
----
\(x = 6 \text{ and } y = 0\)
----
\(x\)
----
\(y\)
----
\[\begin{alignrll}
		y &= \dfrac{5}{x - 2} & \ldots (1)\\[8pt]
		y + 1  &= 2x & \ldots (2)
		\end{alignrll}\]
----
\(y\)
----
\[\begin{align}
		 					y + 1  &= 2x \\
								y  &= 2x - 1
		   				\end{align}\]
----
\(x\)
----
\[\begin{align}
		 					2x - 1 &= \frac{5}{x - 2} \\
		 					(2x - 1)(x - 2) &= 5 \\
		 					2x^2 - 5x + 2 &= 5 \\
		 					2x^2 - 5x - 3 &= 0 \\
		 					(2x + 1)(x - 3) &= 0 \\
		 					\text{Therefore } x = -\frac{1}{2}&\text{ or } x = 3
		   				\end{align}\]
----
\(x\)
----
\(y\)
----
\(x -\dfrac{1}{2}\)
----
\[\begin{align}
		  y &= 2(-\frac{1}{2}) - 1  \\
		  \therefore y &= -2
		  \end{align}\]
----
\((-\dfrac{1}{2};-2)\)
----
\(x = 3\)
----
\[\begin{align}
		  y &= 2(3) -1 \\
		  &= 5
		  \end{align}\]
----
\((3;5)\)
----
\(x = -\dfrac{1}{2} \text{ and } y = -2\)
----
\(x = 3 \text{ and } y = 5\)
----
\(y\)
----
\(x\)
----
\(y\)
----
\[\begin{alignrll}
		    y + x^2 &= 1 & \ldots (1) \\[8pt]
		    y - x + 5 &= 0 & \ldots (2)
		  \end{alignrll}\]
----
\(y\)
----
\[\begin{align}
		  y + x^2 &= 1  \\
		  y  &= - x^2 + 1 
		   \end{align}\]
----
\[\begin{align}
		  y - x + 5 &= 0  \\
		  y &= x - 5  
		   \end{align}\]
----
\((-3;-8)\)
----
\((2;-3)\)
----
\((-3;-8)\)
----
\((2;-3)\)
----
\(y+x=5\)
----
\(y-{x}^{2}+3x-5=0\)
----
\(y\)
----
\[\begin{align}
		 					y + x  &= 5 \\
								y  &= 5 - x
		   				\end{align}\]
----
\[\begin{align}
		 					y - x^{2} + 3x - 5  &= 0 \\
							y  &= x^{2} - 3x + 5
		   				\end{align}\]
----
\(x\)
----
\[\begin{align}
		 					5 - x &= x^{2} - 3x + 5 \\
		 					0 &= x^{2} - 2x  \\
		 					0 &= x(x - 2) \\
		 					\text{Therefore } x = 0 &\text{ or } x = 2
		   				\end{align}\]
----
\(x\)
----
\(y\)
----
\(x = 0\)
----
\[\begin{align}
		  y &= 5 - 0  \\
		  \therefore y &= 5
		  \end{align}\]
----
\((0;5)\)
----
\(x = 2\)
----
\[\begin{align}
		  y &= 5 - 2 \\
		  &= 3
		  \end{align}\]
----
\((2;3)\)
----
\(x = 0 \text{ and } y = 5\)
----
\(x = 2 \text{ and } y = 3\)
----
\((0;5)\)
----
\((2;3)\)
----
\(y = 6 - 5x + x^2\)
----
\(y-x+1=0\)
----
\(y\)
----
\[y  = 6 - 5x + x^2\]
----
\[\begin{align}
		 					y - x + 1  &= 0 \\
							y  &= x - 1
		   				\end{align}\]
----
\(x\)
----
\[\begin{align}
		 					6 - 5x + x^2 &= x - 1 \\
		 					x^{2} - 6x + 7 &= 0 \\
		 					x &= \dfrac{-(-6) \pm \sqrt{(-6)^{2} - 4(1)(7)}}{2(1)} \\
							x &= \dfrac{6 \pm \sqrt{36 - 28}}{2} \\
							x &= \dfrac{6 \pm \sqrt{8}}{2} \\
							x &= \dfrac{6 \pm 2\sqrt{2}}{2} \\
							x &= 3 \pm \sqrt{2}  
		   				\end{align}\]
----
\(x\)
----
\(y\)
----
\(x = 3 - \sqrt{2}\)
----
\[\begin{align}
		  y &= 3 - \sqrt{2} - 1  \\
		  \therefore y &= 2 - \sqrt{2}
		  \end{align}\]
----
\(x = 3 + \sqrt{2}\)
----
\[\begin{align}
		  y &= 3 + \sqrt{2} - 1 \\
		  &= 2 + \sqrt{2}
		  \end{align}\]
----
\(x = 3 \pm \sqrt{2} \text{ and } y = 2 \pm \sqrt{2}\)
----
\(y=\dfrac{2x+2}{4}\)
----
\(y-2{x}^{2}+3x+5=0\)
----
\(y\)
----
\[y = \dfrac{2x+2}{4}\]
----
\[\begin{align}
		 					y - 2{x}^{2} + 3x + 5 &= 0 \\
							y  &= 2x^{2} - 3x - 5
		   				\end{align}\]
----
\(x\)
----
\[\begin{align}
		 					\dfrac{2x+2}{4} &= 2x^{2} - 3x - 5 \\
		 					2x + 2 &= 8x^{2} - 12x - 20 \\
							0 &= 8x^{2} - 14x - 22 \\
							0 &= 4x^{2} - 7x - 11 \\
		 					0 &= (4x - 11)(x + 1) \\
		 					\text{Therefore } x = -1 &\text{ or } x = \frac{11}{4}
		   				\end{align}\]
----
\(x\)
----
\(y\)
----
\(x = -1\)
----
\[\begin{align}
		  y &= \dfrac{2(-1) + 2}{4} \\
		  y &= 0
		  \end{align}\]
----
\((-1;0)\)
----
\(x = \frac{1}{4}\)
----
\[\begin{align}
		  y &= \dfrac{2(\frac{1}{4}) + 2}{4} \\
		  y &= \dfrac{\frac{5}{2}}{4} \\
		  &= \frac{5}{8}
		  \end{align}\]
----
\((\frac{1}{4};\frac{5}{8})\)
----
\(x = -1 \text{ and } y = 0\)
----
\(x = \frac{1}{4} \text{ and } y = \frac{5}{8}\)
----
\((-1;0)\)
----
\((\frac{1}{4};\frac{5}{8})\)
----
\(a - 2b - 3 = 0\)
----
\(a-3{b}^{2}+4=0\)
----
\(a\)
----
\[a = 2b + 3\]
----
\[\begin{align}
		 					a - 3{b}^{2} + 4 &= 0 \\
							a  &= 3b^{2} - 4
		   				\end{align}\]
----
\(b\)
----
\[\begin{align}
		 					2b + 3 &= 3b^{2} - 4 \\
							0 &= 3b^{2} - 2b - 7 \\
		 					0 &= \dfrac{-(-2) \pm \sqrt{(-2)^{2} - 4(3)(-7)}}{2(3)} \\
							b &= \dfrac{2 \pm \sqrt{4 + 84}}{6} \\
							b &= \dfrac{2 \pm \sqrt{88}}{6}
		   				\end{align}\]
----
\(b\)
----
\(a\)
----
\(b = \dfrac{2 + \sqrt{88}}{6}\)
----
\[\begin{align}
		  a &= 2\left(\dfrac{2 + \sqrt{88}}{6}\right) + 3  \\
		  &= \dfrac{2 + \sqrt{88}}{3} + 3 \\
		  &= \dfrac{11 + \sqrt{88}}{3}
		  \end{align}\]
----
\(b = \dfrac{2 - \sqrt{88}}{6}\)
----
\[\begin{align}
		  a &= 2\left(\dfrac{2 - \sqrt{88}}{6}\right) + 3  \\
		  &= \dfrac{2 - \sqrt{88}}{3} + 3 \\
		  &= \dfrac{11 - \sqrt{88}}{3}
		  \end{align}\]
----
\(b = \dfrac{2 \pm \sqrt{88}}{6} \text{ and } a = \dfrac{11 \pm \sqrt{88}}{6}\)
----
\(x^2 - y + 2 = 3x\)
----
\(4x = 8 + y\)
----
\(y\)
----
\[\begin{align}
		 					x^{2} + y + 2 &= 3x \\
							y  &= -x^{2} + 3x - 2
		   				\end{align}\]
----
\[y = 4x - 8\]
----
\(x\)
----
\[\begin{align}
		 					4x - 8 &= -x^{2} + 3x - 2 \\
		 					x^{2} + x - 6 &= 0 \\
							(x + 3)(x - 2) & = 0 \\
		 					\text{Therefore } x = -3 &\text{ or } x = 2
		   				\end{align}\]
----
\(x\)
----
\(y\)
----
\(x = -3\)
----
\[\begin{align}
		  y &= 4(-3) - 8 \\
		  y &= -20
		  \end{align}\]
----
\((-3;-20)\)
----
\(x = 2\)
----
\[\begin{align}
		  y &= 4(2) - 8 \\
		  &= 0
		  \end{align}\]
----
\((2;0)\)
----
\(x = -3 \text{ and } y = -20\)
----
\(x = 2 \text{ and } y = 0\)
----
\((-3;-20)\)
----
\((2;0)\)
----
\(2y + x^2 + 25 = 7x\)
----
\(3x = 6y + 96\)
----
\(y\)
----
\[\begin{align}
		 					2y + x^2 + 25 &= 7x \\
							2y + x^2 + 25 &= 7x - x^{2} + 25 \\
							y  &= \frac{-x^{2} + 7x + 25}{2}
		   				\end{align}\]
----
\[\begin{align}
		 		2y & = x - 32 \\
				y & = \frac{x - 32}{2} 
				\end{align}\]
----
\(x\)
----
\[\begin{align}
		 					\frac{x - 32}{2} &= \frac{-x^{2} + 7x + 25}{2} \\
		 					x - 32 &= -x^{2} + 7x + 25 \\
							x^{2} - 6x - 57 & = 0 \\
							x & = \dfrac{-(-6) \pm \sqrt{(-6)^{2} - 4(1)(-57)}}{2(1)} \\
							& = \dfrac{6 \pm \sqrt{36 + 228}}{2} \\
							& = \dfrac{6 \pm \sqrt{264}}{2}
		   				\end{align}\]
----
\(x\)
----
\(y\)
----
\(x = \dfrac{6 + \sqrt{264}}{2}\)
----
\[\begin{align}
		  2y &= \dfrac{6 + \sqrt{264}}{2} - 32  \\
		  y &= \dfrac{70 + \sqrt{264}}{4} \\
		  &= \dfrac{70 + \sqrt{264}}{4}
		  \end{align}\]
----
\(x = \dfrac{6 - \sqrt{264}}{2}\)
----
\[\begin{align}
		  2y &= \dfrac{6 - \sqrt{264}}{2} - 32  \\
		  y &= \dfrac{70 - \sqrt{264}}{4} \\
		  &= \dfrac{70 - \sqrt{264}}{4}
		  \end{align}\]
----
\(x = \dfrac{6 \pm \sqrt{264}}{2} \text{ and } y = \dfrac{70 \pm \sqrt{264}}{2}\)
----
\({x}^{2}-1-y=0\)
----
\(y+x-5=0\)
----
\(y\)
----
\[\begin{align}
		  x^{2} - 1 - y &= 0  \\
		  y  &= x^2 - 1 
		   \end{align}\]
----
\[\begin{align}
		  y + x - 5 &= 0  \\
		  y &= -x + 5  
		   \end{align}\]
----
\((-3;8)\)
----
\((2;3)\)
----
\((-3;8)\)
----
\((2;3)\)
----
\(x+y-10=0\)
----
\({x}^{2}-2-y=0\)
----
\(y\)
----
\[\begin{align}
		  x + y - 10 &= 0  \\
		  y  &= -x + 10 
		   \end{align}\]
----
\[\begin{align}
		  x^{2} - 2 - y &= 0  \\
		  y &= x^{2} - 2  
		   \end{align}\]
----
\((-4;14)\)
----
\((3;7)\)
----
\((-4;14)\)
----
\((3;7)\)
----
\(xy = 12\)
----
\(7 = x+y\)
----
\(y\)
----
\[\begin{align}
		  xy &= 12  \\
		  y  &= \frac{12}{x} 
		   \end{align}\]
----
\[\begin{align}
		  7 &= x + y  \\
		  y &= 7 - x  
		   \end{align}\]
----
\(x \ne 0\)
----
\(y \ne 0\)
----
\((3;4)\)
----
\((4;3)\)
----
\((3;4)\)
----
\((4;3)\)
----
\(6-4x-y=0\)
----
\(12-2{x}^{2}-y=0\)
----
\(y\)
----
\[\begin{align}
		  6 - 4x - y &= 0  \\
		  y  &= -4x + 6 
		   \end{align}\]
----
\[\begin{align}
		  12 - 2x^{2} - y &= 0  \\
		  y &= -2x^{2} + 12 
		   \end{align}\]
----
\((4;-10)\)
----
\((-2;14)\)
----
\[\begin{align}
					-4x + 6 &= -2x^{2} + 12 \\
					2x - 3  &= x^{2} - 6 \\
					x^{2} - 2x  - 9 & = 0 \\
					x & = \dfrac{-(-2) \pm \sqrt{(-2)^{2} - 4(1)(-9)}}{2(1)} \\
					x & = \dfrac{2 \pm \sqrt{(4 + 36}}{2} \\
					x & = \dfrac{2 \pm \sqrt{40}}{2}
					\end{align}\]
----
\(y\)
----
\[\begin{align}
					y & = -4\left(\dfrac{2 \pm \sqrt{40}}{2} \right) + 6 \\
					&= -2(2 \pm \sqrt{40}) + 6 \\
					& = -4 \pm 2\sqrt{40} + 6 \\
					& = 2 \pm 2\sqrt{40}
					\end{align}\]
----
\(x\)
----
\(y\)
----
\(x\)
----
\(y\)
----
\(\dfrac{5}{7}\)
----
\(x\)
----
\(x\)
----
\(x\)
----
\(y\)
----
\[\begin{alignrll}
		  4x + 3y &= 54 & \ldots (1) \\[8pt]
		  x &= y + 3 & \ldots (2)
		\end{alignrll}\]
----
\[\begin{align}
		  4(y+3) + 3y &= 54 \\
		  4y+12 + 3y &= 54 \\
		  7y &= 42 \\
		  y &= 6
		   \end{align}\]
----
\(y=6\)
----
\[\begin{align}

		  		x &= y + 3 \\
		    	&= 6 + 3 \\
    			&= 9
		   \end{align}\]
----
\(x\)
----
\(x-3\)
----
\[4x + 3(x-3) = 54\]
----
\(x\)
----
\[\begin{align}
		  4x + 3(x-3) &= 54 \\
		  4x + 3x-9 &= 54 \\
		  7x &= 63 \\
		  x &= 9
		   \end{align}\]
----
\(x\)
----
\(x - 2\)
----
\[\begin{align}
		  0 &= 4x^2 - 23x + 15 \\
			0 &= (4x-3)(x-5) 
		   \end{align}\]
----
\(x = \frac{3}{4}\)
----
\(x = 5\)
----
\(\frac{3}{4}\)
----
\(x = \dfrac{3}{4}\)
----
\(l\)
----
\(b\)
----
\(A = l \times b\)
----
\(l = 2b\)
----
\[\begin{align}
				A = 8 & = l \times b \\
				& = 2b(b) \\
				& = 2b^{2} \\
				b^{2} & = 4 \\
				b & = \pm 2
				\end{align}\]
----
\(a\)
----
\(b\)
----
\(c\)
----
\(d\)
----
\[\begin{align}
				c & = 80 + b \\
				a & = c - 110 \\
				a + b & = 208 \\
				\frac{a + b + c + d}{4} & = 146
				\end{align}\]
----
\(c\)
----
\[\begin{align}
				c & = 80 + b \\
				c & = a + 110 
				\end{align}\]
----
\(a = 208 - b\)
----
\(b\)
----
\[\begin{align}
				80 + b & = 208 - b + 110 \\
				2b & = 208 + 110 - 80 \\
				2b & = 238 \\
				b & = 119
				\end{align}\]
----
\(a\)
----
\[\begin{align}
				a + b & = 208 \\
				a + 119 & = 208 \\
				a & = 89
				\end{align}\]
----
\(c\)
----
\[\begin{align}
				c & = 80 + b \\
				c & = 80 + 208 \\
				c & = 288
				\end{align}\]
----
\(d\)
----
\[\begin{align}
				\frac{a + b + c + d}{4} & = 164 \\
				496 + d & = 656 \\
				d & = 187
				\end{align}\]
----
\(d\)
----
\(t\)
----
\(s=5{t}^{2}+{v}_{0}t\)
----
\({v}_{0}\)
----
\(t\)
----
\[\begin{align}
				  s & = 5{t}^{2} + {v}_{0}t \\
				  500 & = 5t^{2} + 5t \\
				  t^{2} + t - 100 & = 0 \\
				  t & = \dfrac{-(1) \pm \sqrt{(1)^{2} - 4(1)(100)}}{2(1)} \\
				  & = \dfrac{1 \pm \sqrt{1 + 400}}{2} \\
				  & = \dfrac{1 \pm \sqrt{401}}{2}
				  \end{align}\]
----
\(t = 5d\)
----
\(t = 5(21) = 105\)
----
\(y\)
----
\(t = 5d\)
----
\(P\)
----
\(I\)
----
\[\begin{align}
					  P & = 12(10) - \frac{1}{2}(10)^{2} \\
					  & = 120 - 50 \\
					  & = 70
					  \end{align}\]
----
\(3x\)
----
\(4x\)
----
\(5x\)
----
\(y\)
----
\(y=\frac{300-{x}^{2}}{x}\)
----
\(a \times b = 0\)
----
\(a = 0\)
----
\(b=0\)
----
\(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
----
\(\Delta = b^2 - 4ac\)
----
\(\Delta <0\)
----
\(\Delta = 0\)
----
\(\Delta > 0\)
----
\(Δ =\)
----
\(Δ =\)
----
\({x}^{2}-x-1=0\)
----
\(16\left(x+1\right)={x}^{2}\left(x+1\right)\)
----
\[\begin{align}
  		16(x+1) &=x^2(x+1)\\
		16(x+1) - x^2(x+1) &= 0\\
		(16-x^2)(x+1)&=0\\
		x^2= 16& \text{ or } x=-1\\
		x  = \pm 4& \text{ or } x=-1
		\end{align}\]
----
\(x=\pm 4\)
----
\({y}^{2}+3+\dfrac{12}{{y}^{2}+3}=7\)
----
\[\begin{align}
		\text{Let } y^2+3&=k\\
		\text{Restriction: } y^2+3&\ne 0 \\
		\text{Therefore } k &\ne 0 \\
		\therefore k + \frac{12}{k} &= 7\\
		k^2 + 12 &= 7k\\
		k^2 - 7k + 12 &= 0\\
		(k-3)(k-4) &= 0\\
		k=3 &\text{ or } k=4\\
		\therefore y^2 + 3 =3 &\text{ or } y^2 + 3 = 4\\
		\therefore y^2 =0 &\text{ or }  y^2 = 1\\
		\therefore y=0 &\text{ or } y= \pm 1
		\end{align}\]
----
\(y=0\)
----
\(y = \pm 1\)
----
\(x\)
----
\(2{x}^{4}-5{x}^{2}-12=0\)
----
\[\begin{align}
 		 2x^4 - 5x^2 - 12 &= 0\\
		(x^2-4)(2x^2+3)&=0\\
		x^2 = 4 & \text{ or } x^2 = -\frac{3}{2} \quad \text{(no real solution)}\\
		\therefore x &= \pm 2
		\end{align}\]
----
\(x= \pm 2\)
----
\(x\)
----
\(x\left(x-9\right)+14=0\)
----
\[\begin{align}
      	x(x-9) + 14 &= 0\\
    	x^2 - 9x + 14 &= 0\\
    	(x-7)(x-2)&=0\\
    	x = 7 &\text{ or } x=2
	\end{align}\]
----
\(x=7\)
----
\(x = 2\)
----
\(x^2 - x = 3\)
----
\(x + 2 = \frac{6}{x}\)
----
\(\dfrac{1}{x+1}+\dfrac{2x}{x-1}=1\)
----
\[\begin{align}
    	\dfrac{1}{x+1} + \dfrac{2x}{x-1} &= 1\\
    	(x-1) + 2x(x+1) &= (x+1)(x-1)\\
    	x-1 + 2x^2 + 2x &= x^2 - 1\\
    	x^2 + 3x &= 0\\
    	x(x+3)&=0\\
   	x=0& \text{ or } x=-3 
	\end{align}\]
----
\(x=0\)
----
\(x = -3\)
----
\(x\)
----
\(p\)
----
\({x}^{2}-px-4=0\)
----
\[\begin{align}
				x^{2} - px - 4 & = 0 \\
				x^{2} - px + \frac{p^{2}}{4} & = 4 + \frac{p^{2}}{4} \\
				\left(x - \frac{p}{2} \right)^{2} & = \frac{16 + p^{2}}{4} \\
				x - \frac{p}{2} & = \sqrt{\frac{16 + p^{2}}{4}} \\
				x & = \frac{\sqrt{16+p^{2}}}{2} - \frac{p}{2} \\
				& = \frac{\sqrt{16 + p^{2}} - 2}{2}
				\end{align}\]
----
\(x=\frac{\sqrt{16 + p^{2}} - 2}{2}\)
----
\(a{x}^{2}+bx+c=0\)
----
\(x=\dfrac{2}{3}\)
----
\(x=-4\)
----
\(a\)
----
\(b\)
----
\(c\)
----
\[\begin{align}
  		(3x- 2)(x+4)&=0\\
		3x^2 + 12x - 2x - 8 &= 0\\
		3x^2 + 10x - 8 &= 0\\
		\therefore a &= 3\\
		b &= 10\\
		c&= -8
		\end{align}\]
----
\(a = 3\)
----
\(b=10\)
----
\(c=-8\)
----
\(4{x}^{2}+px-9=0\)
----
\(p\)
----
\[4x^2 + px - 9 = 0\]
----
\[\begin{align}
		x &= \frac{-p \pm \sqrt{p^2 - 4(4)(-9)}}{8} \\
 		&= \pm \dfrac{\sqrt{144 + p^2}}{8}
	      \end{align}\]
----
\[\begin{align}
		\therefore x_1 = \dfrac{-p + \sqrt{144 + p^2}}{8} & \text{ and } 
		x_2 = \dfrac{-p - \sqrt{144 + p^2}}{8}\\
		x_1 - x_2 &= 5\\
		\therefore \dfrac{-p + \sqrt{144 + p^2}}{8} - \dfrac{-p - \sqrt{144 + p^2}}{8} &= 5\\
		-p + \sqrt{144 + p^2} + p + \sqrt{144 + p^2} &= 40\\
		\sqrt{p^2 + 144} &= 20\\
		p^2 + 144 &= 400\\
		p^2 &= 256\\
		\therefore p &= \pm 16
		\end{align}\]
----
\(p = \pm 16\)
----
\({x}^{2}+bx+c=0\)
----
\(x\)
----
\[\begin{align}
		 \text{Saskia:}\\
		(x+4)(x-2)&=0\\
		x^2 + 2x - 8  &= 0\\
		\therefore a = 1 &  \text{ and } b = 2\\
		\\
 		\text{Sven:}\\
		(x-1)(x+15) &= 0\\
		x^2 + 14x -15 &= 0\\
		\therefore c &= -15\\
		\\
 		\text{Correct equation:}\\
		x^2 + 2x-15 &=0\\
		(x+5)(x-3) &= 0\\
		\therefore   \text{correct roots are } x = -5&  \text{ and } x=3
		\end{align}\]
----
\(x^{2} + 2x - 15\)
----
\(b\)
----
\(\dfrac{b^2 - 5b + 6}{b + 2}\)
----
\(b + 2 = 0\)
----
\(b = -2\)
----
\[\begin{align}
			  \dfrac{b^2 - 5b + 6}{b + 2} & = 0 \\
			  \dfrac{(b - 2)(b - 3)}{b + 2} & = 0
			  \end{align}\]
----
\(b = 2\)
----
\(b = 3\)
----
\(b=-2\)
----
\(b=-2\)
----
\(b = 2\)
----
\(b = 3\)
----
\(\dfrac{(x^2 - 6)(2x + 1)}{x + 2} = 0\)
----
\(x\)
----
\(x\)
----
\(x\)
----
\(x\)
----
\(x\)
----
\(x \ne -2\)
----
\[\begin{align}
				(x^2 - 6)(2x + 1) & = 0 \\
				x = \frac{-1}{2} & \text{ or } x = \pm \sqrt{6}
				\end{align}\]
----
\(x = \frac{-1}{2}\)
----
\(x = \pm \sqrt{6}\)
----
\(\dfrac{(x-6)^{\frac{1}{2}}}{x^2+3}\)
----
\(x\)
----
\(x = 6\)
----
\(x\)
----
\(x = \pm \sqrt{3}\)
----
\(a\)
----
\(\dfrac{\sqrt{8-2a}}{a-3} \geq 0\)
----
\(a \ne 3\)
----
\[\begin{align}
				\sqrt{8 - 2a} & = 0 \\
				8 - 2a & = 0 \\
				a & = 4 
				\end{align}\]
----
\(a=3\)
----
\(a=4\)
----
\(a-3\)
----
\(-\)
----
\(+\)
----
\(+\)
----
\(+\)
----
\(a-4\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(+\)
----
\(-\)
----
\(-\)
----
\(+\)
----
\(a \geq 4\)
----
\(a{x}^{2}+bx+c=0\)
----
\[x = \frac{2c}{-b \pm \sqrt{b^2 - 4ac}}\]
----
\(2x^2 + x - 3 = 0\)
----
\[\begin{alignrll}
  		ax^2 + bx + c &= 0 & \\[8pt]
		a + \dfrac{b}{x} + \dfrac{c}{x^2} &= 0 & \text{Divided by } x^2 \text{ where } x\ne 0 \\[8pt]
		\dfrac{c}{x^2} + \dfrac{b}{x} + a &= 0 & \text{Rearranged} \\[8pt]
		\dfrac{1}{x^2} + \dfrac{b}{cx} + \dfrac{a}{c} &= 0 & \text{Divided by } c \text{ where } c \ne 0 \\[8pt]
		\dfrac{1}{x^2} + \dfrac{b}{cx} &= -\dfrac{a}{c} & \text{Subtracted } \frac{a}{c} \text{ from both sides} \\[8pt]
		\therefore \dfrac{1}{x^2} + \dfrac{b}{cx} + \dots & & \text{Got stuck}
		\end{alignrll}\]
----
\[\begin{align}
  		x =& \dfrac{2c}{-b \pm \sqrt{b^2 - 4ac}}\\
		&= \dfrac{2(-3)}{-(1) \pm \sqrt{1^2 - 4(2)(-3)}}\\
		&=\dfrac{-6}{-1 \pm \sqrt{25}}\\
		&= \dfrac{-6}{-1 \pm 5}\\
		x = \frac{-6}{4} = -\frac{3}{2} & \text{ or } x = \frac{-6}{-6} = 1
		\end{align}\]
----
\[\begin{align}
  		2x^2 + x -3 &= 0\\
		(2x+3)(x-1)&=0\\
		x = -\frac{3}{2}& \text{ or } x=1
		\end{align}\]
----
\[\begin{align}
  		ax^2 + bx + c &= 0 \\
		a + \dfrac{b}{x} + \dfrac{c}{x^2} &= 0 \\
		\dfrac{c}{x^2} + \dfrac{b}{x} + a &= 0 \\
		\dfrac{1}{x^2} + \dfrac{b}{cx} + \dfrac{a}{c} &= 0 \\
		\dfrac{1}{x^2} + \dfrac{b}{cx} &= -\dfrac{a}{c} \\
		\dfrac{1}{x^2} + \dfrac{b}{cx} + \dfrac{b^2}{4c^2} &= -\dfrac{a}{c} +\dfrac{b^2}{4c^2}\\
		\left(\dfrac{1}{x} + \dfrac{b}{2c}\right)^2 &= \dfrac{-4ac +b^2}{4c^2}  = \dfrac{b^2-4ac}{4c^2}\\
		\dfrac{1}{x} + \dfrac{b}{2c} &= \pm \sqrt{\dfrac{b^2-4ac}{4c^2}}\\
		\dfrac{1}{x} &= - \dfrac{b}{2c} \pm \sqrt{\dfrac{b^2-4ac}{4c^2}}\\
		\dfrac{1}{x} &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2c}\\
		\therefore x &= \dfrac{2c}{-b \pm \sqrt{b^2 - 4ac}}
		\end{align}\]
----
\(x = -\frac{3}{2} \text{ or } x=1\)
----
\(x\)
----
\(\dfrac{4}{x-3} \le 1\)
----
\(x < 3 \text{ or } x \geq 7\)
----
\(\dfrac{4}{{\left(x-3\right)}^{2}}<1\)
----
\(x < 1 \text{ or } x > 5\)
----
\(\dfrac{2x-2}{x-3}>3\)
----
\(3 < x < 7\)
----
\(\dfrac{-3}{\left(x-3\right)\left(x+1\right)}<0\)
----
\(x < -1\)
----
\(x > 3\)
----
\({\left(2x-3\right)}^{2}<4\)
----
\(2x\le \dfrac{15-x}{x}\)
----
\(x \leq -3 \text{ or } 0 < x \leq \frac{5}{2}\)
----
\(\dfrac{{x}^{2}+3}{3x-2}\le 0\)
----
\(x < \frac{2}{3}\)
----
\(x-2\ge \dfrac{3}{x}\)
----
\(-1 \leq x < 0\)
----
\(x \geq 3\)
----
\(\dfrac{{x}^{2}+3x-4}{5+{x}^{4}}\le 0\)
----
\(-4 \le x \le 1\)
----
\(\dfrac{x-2}{3-x}\ge 1\)
----
\(2\frac{1}{2} \le x < 3\)
----
\(y-2x=0\)
----
\(y-{x}^{2}-2x+3=0\)
----
\(y\)
----
\[\begin{align}
		 					y - 2x  &= 0 \\
							y  &= 2x
		   				\end{align}\]
----
\[\begin{align}
		 					y - x^{2} - 2x + 3  &= 0 \\
							y  &= x^{2} + 2x - 3
		   				\end{align}\]
----
\(x\)
----
\[\begin{align}
		 					2x &= x^{2} + 2x - 3 \\
		 					0 &= x^{2} -3  \\
							x^{2} & = 3 \\
		 					x &= \pm \sqrt{3}
		   				\end{align}\]
----
\(x\)
----
\(y\)
----
\(x = \sqrt{3}\)
----
\[y = 2\sqrt{3}\]
----
\((\sqrt{3};2\sqrt{3})\)
----
\(x = -\sqrt{3}\)
----
\[y = -2(\sqrt{3})\]
----
\((-\sqrt{3};-2\sqrt{3})\)
----
\(x = \pm \sqrt{3} \text{ and } y = \pm 2\sqrt{3}\)
----
\(a-3b=0\)
----
\(a-{b}^{2}+4=0\)
----
\(a\)
----
\[\begin{align}
		 					a - 3b  &= 0 \\
							a  &= 3b
		   				\end{align}\]
----
\[\begin{align}
		 					a - b^{2} + 4  &= 0 \\
							a  &= b^{2} - 4
		   				\end{align}\]
----
\(b\)
----
\[\begin{align}
		 					3b &= b^{2} - 4 \\
		 					0 &= b^{2} - 3b - 4  \\
							(b - 4)(b + 1) & = 0 \\
		 					b = 4 & \text{ or } b = -1
		   				\end{align}\]
----
\(b\)
----
\(a\)
----
\(b = 4\)
----
\[a = 12\]
----
\(b = -1\)
----
\[a = -3\]
----
\(a = -3 \text{ and } b = -1\)
----
\(a = 12 \text{ and } b = 4\)
----
\(y-{x}^{2}-5x=0\)
----
\(10 = y - 2x\)
----
\(y\)
----
\[\begin{align}
		 					y - x^{2} - 5x  &= 0 \\
							y  &= x^{2} + 5x
		   				\end{align}\]
----
\[\begin{align}
		 					10 &= y - 2x \\
							y  &= 2x + 10
		   				\end{align}\]
----
\(x\)
----
\[\begin{align}
		 					x^{2} + 5x &= 2x + 10 \\
		 					x^{2} + 3x - 10 &= 0 \\
							(x + 5)(x - 2) & = 0 \\
		 					x = -5 & \text{ or } x = 2
		   				\end{align}\]
----
\(x\)
----
\(y\)
----
\(x = -5\)
----
\[\begin{align}
			y & = 2(-5) + 10 \\
			& = 0
		  \end{align}\]
----
\(x = 2\)
----
\[\begin{align}
			y & = 2(2) + 10 \\
			& = 14
		  \end{align}\]
----
\(x = -5 \text{ and } y = 0\)
----
\(x = 2 \text{ and } y = 14\)
----
\(p = 2p^2 + q - 3\)
----
\(p - 3q = 1\)
----
\(q\)
----
\[\begin{align}
		 					p &= 2p^{2} + q - 3 \\
							q &= -2p^{2} + p + 3 
		   				\end{align}\]
----
\[\begin{align}
		 					p - 3q &= 1 \\
							q &= \frac{p - 1}{3}
		   				\end{align}\]
----
\(p\)
----
\[\begin{align}
		 					-2p^{2} + p + 3 &= \frac{p - 1}{3} \\
		 					-6p^{2} + 3p + 9 &= p - 1 \\
							6p^{2} - 4p - 10 &= 0 \\
							3p^{2} - 2p - 5 &= 0 \\
							(3p - 5)(p + 1) & = 0 \\
		 					p = \frac{5}{3} & \text{ or } p = -1
		   				\end{align}\]
----
\(p\)
----
\(q\)
----
\(p = \frac{5}{3}\)
----
\[\begin{align}
			q & = \frac{\frac{5}{3} - 1}{3} \\
			& = \frac{2}{9}
		  \end{align}\]
----
\(p = -1\)
----
\[\begin{align}
			q & = \frac{(-1) - 1}{3} \\
			& = \frac{-2}{3}
		  \end{align}\]
----
\(p = \frac{5}{3} \text{ and } q = \frac{2}{9}\)
----
\(p = -1 \text{ and } q = \frac{-2}{3}\)
----
\(a-{b}^{2}=0\)
----
\(a-3b+1=0\)
----
\(a\)
----
\[\begin{align}
		 					a - b^{2} &= 0 \\
							a &= b^{2} 
		   				\end{align}\]
----
\[\begin{align}
		 					a - 3b + 1 &= 0 \\
							a &= 3b - 1
		   				\end{align}\]
----
\(b\)
----
\[\begin{align}
		 					b^{2} &= 3b - 1 \\
		 					b^{2} - 3b + 1 &= 0 \\
							b & = \frac{-(-3) \pm \sqrt{(-3)^{2} - 4(1)(1)}}{2(1)} \\
							& = \frac{3 \pm \sqrt{5}}{2}
		   				\end{align}\]
----
\(b\)
----
\(a\)
----
\(b = \frac{3 + \sqrt{5}}{2}\)
----
\[\begin{align}
			a & = 3\left(\frac{3 + \sqrt{5}}{2} \right) - 1 \\
			& = \frac{7 + 3\sqrt{5}}{2}
		  \end{align}\]
----
\(b = \frac{3 - \sqrt{5}}{2}\)
----
\[\begin{align}
			a & = 3\left(\frac{3 - \sqrt{5}}{2} \right) - 1 \\
			& = \frac{7 - 3\sqrt{5}}{2}
		  \end{align}\]
----
\(b = \frac{3 \pm \sqrt{5}}{2} \text{ and } a = \frac{7 \pm 3\sqrt{5}}{2}\)
----
\(a-2b+1=0\)
----
\(a-2{b}^{2}-12b+4=0\)
----
\(a\)
----
\[\begin{align}
		 					a - 2b + 1 &= 0 \\
							a &= 2b - 1 
		   				\end{align}\]
----
\[\begin{align}
		 					a - 2{b}^{2} - 12b + 4 &= 0 \\
							a &= 2{b}^{2} + 12b - 4
		   				\end{align}\]
----
\(b\)
----
\[\begin{align}
		 					2b - 1 &= 2{b}^{2} + 12b - 4 \\
		 					2b^{2} + 10b - 5 &= 0 \\
							b & = \frac{-(10) \pm \sqrt{(10)^{2} - 4(2)(-5)}}{2(2)} \\
							& = \frac{-10 \pm \sqrt{140}}{4}
		   				\end{align}\]
----
\(b\)
----
\(a\)
----
\(b = \frac{-10 + \sqrt{140}}{4}\)
----
\[\begin{align}
			a & = 2\left(\frac{-10 + \sqrt{140}}{4} \right) - 1 \\
			& = \frac{-24 + 2\sqrt{140}}{4} \\
			& = \frac{-12 + \sqrt{140}}{2}
		  \end{align}\]
----
\(b = \frac{-10 - \sqrt{140}}{4}\)
----
\[\begin{align}
			a & = 2\left(\frac{-10 - \sqrt{140}}{4} \right) - 1 \\
			& = \frac{-24 - 2\sqrt{140}}{4} \\
			& = \frac{-12 - \sqrt{140}}{2}
		  \end{align}\]
----
\(b = \frac{-10 \pm \sqrt{140}}{4} \text{ and } a = \frac{-12 \pm \sqrt{140}}{2}\)
----
\(y+4x-19=0\)
----
\(8y+5{x}^{2}-101=0\)
----
\(y\)
----
\[\begin{align}
		 					y + 4x - 19 &= 0 \\
							y &= -4x + 19 
		   				\end{align}\]
----
\[\begin{align}
		 					8y + 5{x}^{2} - 101 &= 0 \\
							8y &=  -5{x}^{2} + 101 \\
							y &= \frac{-5{x}^{2} + 101}{8}
		   				\end{align}\]
----
\(x\)
----
\(x\)
----
\(y\)
----
\(x = 3\)
----
\[\begin{align}
			y & = -4(3) + 19 \\
			& = 5
		  \end{align}\]
----
\(x = 3 \text{ and } y = 5\)
----
\(a+4b-18=0\)
----
\(2a+5{b}^{2}-57=0\)
----
\(a\)
----
\[\begin{align}
		 					a + 4b - 18 &= 0 \\
							a &= -4b + 18 
		   				\end{align}\]
----
\[\begin{align}
		 					2a + 5{b}^{2} - 57 &= 0 \\
							2a &=  -5b^{2} + 57 \\
							a &= \frac{-5b^{2} + 57}{2}
		   				\end{align}\]
----
\(a\)
----
\(b\)
----
\(a\)
----
\(b = 3 \text{ and } a = 6\)
----
\(2y+x-2=0\)
----
\(8y+{x}^{2}-8=0\)
----
\(y+3x-6=0\)
----
\(y={x}^{2}+4-4x\)
----
\(t\)
----
\(h\left(t\right)=35-5{t}^{2}+30t\)
----
\(t=0\)
----
\(x\)
----
\[P(x) = \frac{55}{2x} + \frac{x}{200} \quad \text{litres per kilometre}\]
----
\(C\)
----
\(-4{y}^{2}-41y-45=0\)
----
\[\begin{align}
-4y^2 - 41y - 45 &= 0\\
4y^2 +41y +45 &= 0\\
(4y+5)(y+9) &= 0\\
y = -\frac{5}{4} &\text{ or } y= -9
\end{align}\]
----
\(y = -\frac{5}{4}\)
----
\(y= -9\)
----
\(16{x}^{2}+20x=36\)
----
\[\begin{align}
16x^2 + 20x - 36 &= 0\\
4x^2 +5x-9 &= 0\\
(4x+9)(x-1)&=0\\
x = -\frac{9}{4} &\text{ or } x=1
\end{align}\]
----
\(x = -\frac{9}{4}\)
----
\(x= 1\)
----
\(42{p}^{2}+104p+64=0\)
----
\[\begin{align}
42p^2 + 104p + 64 &= 0\\
21p^2 + 52p+32&=0\\
(7p+8)(3p+4)&=0\\
p = -\frac{8}{7} &\text{ or } p = -\frac{4}{3}
\end{align}\]
----
\(p = -\frac{8}{7}\)
----
\(p= -\frac{4}{3}\)
----
\(21y+3=54{y}^{2}\)
----
\[\begin{align}
-54y^2 + 21y + 3 &= 0\\
18y^2 - 7y-1 &= 0\\
(9y+1)(2y-1) &= 0\\
y= -\frac{1}{9} &\text{ or } y=\frac{1}{2}
\end{align}\]
----
\(y = -\frac{1}{4}\)
----
\(y= \frac{1}{2}\)
----
\(36{y}^{2}+44y+8=0\)
----
\[\begin{align}
36y^2 + 44y + 8 &= 0\\
9y^2 + 11y+2&=0\\
(9y+2)(y+1)&=0\\
y=-\frac{2}{9} &\text{ or } y=-1
\end{align}\]
----
\(y = -\frac{2}{9}\)
----
\(y= -1\)
----
\(12{y}^{2}-14=22y\)
----
\[\begin{align}
12y^2 - 22y - 14 &= 0\\
6y^2 - 11y - 7 &= 0\\
(3y-7)(2y+1)&=0\\
y= \frac{7}{3} &\text{ or } y=-\frac{1}{2}
\end{align}\]
----
\(y = \frac{7}{3}\)
----
\(y= -\frac{1}{2}\)
----
\(16{y}^{2}+0y-81=0\)
----
\[\begin{align}
16y^2 + 0y - 81 &= 0\\
(4y-9)(4y+9)&=0\\
y=\frac{9}{4} &\text{ or } y=-\frac{9}{4}
\end{align}\]
----
\(y = \frac{9}{4}\)
----
\(y= -\frac{9}{4}\)
----
\(3{y}^{2}+10y-48=0\)
----
\[\begin{align}
3y^2 + 10y - 48 &= 0\\
(3y-8)(y+6)&=0\\
y = \frac{8}{3} &\text{ or } y=-6
\end{align}\]
----
\(y = \frac{8}{3}\)
----
\(y= -6\)
----
\(63-5{y}^{2}=26y\)
----
\[\begin{align}
-5y^2 - 26y + 63 &= 0\\
5y^2+26y-63&=0\\
(5y-9)(y+7) &= 0\\
y = \frac{9}{5} &\text{ or } y=-7
\end{align}\]
----
\(y = \frac{9}{5}\)
----
\(y= -7\)
----
\(2{x}^{2}-30=2\)
----
\[\begin{align}
2x^2-30&=2\\
2x^2-32 &= 0\\
x^2-16&=0\\
(x-4)(x+4)&=0\\
x&= \pm 4
\end{align}\]
----
\(x = \pm 4\)
----
\(2{y}^{2}=98\)
----
\[\begin{align}
2y^2-98&=0\\
y^2-49&=0\\
(y-7)(y+7)&=0\\
y&=\pm 7
\end{align}\]
----
\(y = \pm 7\)
----
\(9y^2 + 32 = ky\)
----
\(k\)
----
\(9y^2 - ky + 32 = 0\)
----
\(k\)
----
\[\begin{align}
				9(8)^{2} - 8k + 32 & = 0 \\
				576 - 8k + 32 & = 0 \\
				8k & = 608 \\
				k & = 76 
				\end{align}\]
----
\[\begin{align}
				9y^{2} - 76y + 32 & = 0 \\
				(9y - 4)(y - 8) & = 0 \\
				y = \frac{4}{9} & \text{ or } y = 8
				\end{align}\]
----
\(k=76\)
----
\(\frac{4}{9}\)
----
\(x\)
----
\(x^2 - x = 6\)
----
\(y\)
----
\((y^2 - y)^2 - (y^2 - y) - 6  = 0\)
----
\[\begin{align}
							x^{2} - x - 6 & = 0 \\
							(x- 3)(x + 2) & = 0 \\
							x = 3 & \text{ or } x = -2 
							\end{align}\]
----
\(y^{2} - y\)
----
\(x = y^{2} - y\)
----
\(x^2 - x = 6\)
----
\(y\)
----
\[\begin{align}
							y^{2} - y & = 3 \\
							y^{2} - y - 3 & = 0 \\
							y & = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(1)(-3)}}{2(1)} \\
							& = \frac{1 \pm \sqrt{13}}{2}
							\end{align}\]
----
\[\begin{align}
							y^{2} - y & = -2 \\
							y^{2} - y + 2 & = 0 \\
							y & = \frac{-(-1) \pm \sqrt{(-1)^{2} - 4(1)(2)}}{2(1)} \\
							& = \frac{1 \pm \sqrt{-7}}{2}
							\end{align}\]
----
\(x\)
----
\(y\)
----
\(x = 3 \text{ or } x = -2   \text{ and } y = \frac{1 \pm \sqrt{13}}{2}\)
----
\(x\)
----
\(x = \sqrt{8-x} + 2\)
----
\[\begin{align}
							x & = \sqrt{ 8 - x} + 2 \\
							x - 2 & = \sqrt{ 8 - x} \\
							(x - 2)^{2} & = 8 - x \\
							x^{2} - 4x + 4 & = 8 - x \\
							x^{2} - 3x - 4 & = 8 - x \\
							(x - 4)(x + 1) & = 0 \\
							x = 4 & \text{ or } x = -1 
							\end{align}\]
----
\(x = 4  \text{ or } x = -1\)
----
\(y\)
----
\(-4y^2 + 8y - 3 = 0\)
----
\(p\)
----
\(4(p-3)^2 - 8(p - 3) + 3 = 0\)
----
\[\begin{align}
							4y^2 - 8y + 3 & = 0 \\
							(2y - 3)(2y - 1) & = 0 \\
							y = \frac{3}{2} & \text{ or } y = \frac{1}{2} 
							\end{align}\]
----
\((p - 3)\)
----
\(y = p - 3\)
----
\(4y^{2} - 8y - 3 = 0\)
----
\(p\)
----
\[\begin{align}
							p - 3 & = \frac{3}{2} \\
							p & = \frac{3}{2} + 3 \\
							p & = \frac{9}{2}
							\end{align}\]
----
\[\begin{align}
							p - 3 & = \frac{1}{2} \\
							p & = \frac{1}{2} + 3 \\
							p & =  \frac{7}{2} 
							\end{align}\]
----
\(y = \frac{3}{2}, y = \frac{1}{2} \text{ and } p = \frac{9}{2}, p = \frac{7}{2}\)
----
\(x\)
----
\(2(x + 3)^{\frac{1}{2}} = 9\)
----
\(x + \dfrac{1}{x} = 3\)
----
\(x^2 + \dfrac{1}{x^2}\)
----
\(x + \dfrac{1}{x} = 3\)
----
\(7\)
----
\(y\)
----
\(5(y-1)^2 - 5 = 19 - (y-1)^2\)
----
\((y-1)^{2}\)
----
\(k\)
----
\[\begin{align}
							5k - 5 & = 19 - k \\
							6k & = 24 \\
							k & = 4 
							\end{align}\]
----
\(y\)
----
\(t\)
----
\(2t(t - \dfrac{3}{2}) = \dfrac{3}{2t^2 - 3t} + 2\)
----
\(t = \frac{1}{2}, t = 1 \text{ or } t = \frac{3 \pm \sqrt{33}}{4}\)
----
\((d)\)
----
\(n\)
----
\(T_1\)
----
\(T_4\)
----
\(T_n\)
----
\(1; 4; 7; 10; 13; \ldots\)
----
\(T_n = 3n-2\)
----
\(n\)
----
\[\begin{align}
  T_1 &= 3(1) - 2 = 1 \\
  T_2 &= 3(2) - 2 = 4 \\
  T_3 &= 3(3) - 2 = 7 \\
  T_4 &= 3(4) - 2 = 10 \\
  T_5 &= 3(5) - 2 = 13
\end{align}\]
----
\(d\)
----
\(d={T}_{2}-{T}_{1}\)
----
\({T}_{1}-{T}_{2}\)
----
\(d\)
----
\[10; 7; 4; 1; \ldots\]
----
\[d = T_n - T_{n-1}\]
----
\[\begin{align}
		  \text{Therefore } d &= T_2 - T_1 \\
		  &= 7-10 \\
		  &= -3 \\
		  \text{or  } d &= T_3 - T_2 \\
		  &= 4-7 \\
		  &= -3 \\
		  \text{or  } d &= T_4 - T_3 \\
		  &= 1-4 \\
		  &= -3
		\end{align}\]
----
\(T_n\)
----
\(d = -3\)
----
\(n = 1\)
----
\[\begin{align}
		  T_1  &= 10 \\
		  &= 10 + (0)(-3) \\
		  &= 10 + (1-1)(-3)
		  \end{align}\]
----
\[\begin{align}
		  T_2  &= 7 \\
		  &= 10 + (1)(-3) \\
		  &= 10 + (2-1)(-3)
		  \end{align}\]
----
\[\begin{align}
		  T_n &= 10 + (n-1)(-3) \\
		  &= 10 -3n + 3\\
		  &= -3n + 13
		  \end{align}\]
----
\(n\)
----
\(45; 29; 13; -3; \ldots\)
----
\(-19;-35;-51\)
----
\(-4; -9; -14; \ldots; -24\)
----
\(T_n = 1-5n\)
----
\(6; \ldots; 24; \ldots; 42\)
----
\(T_n = 9n-3\)
----
\[\begin{aligned}
	T_{2} &= 9(2) - 3 \\
	 &= 18 - 3  \\
	&=  15
	\end{aligned}\]
----
\[\begin{aligned}
	T_{4} &= 9(4) - 3  \\
	 &= 36 - 3  \\
	&= 33
	\end{aligned}\]
----
\(T_2 = 15\)
----
\(T_4 = 33\)
----
\(T_{10}\)
----
\(T_{15}\)
----
\(T_{30}\)
----
\(13; 16; 19; 22; \ldots\)
----
\[\begin{aligned}
	d &= T_2 -  T_1  \\
	 &= 16 - 13  \\
	&= 3 \\
	\therefore T_n &= 10 + 3n
	\end{aligned}\]
----
\[\begin{aligned}
	T_n &= 10 + 3n  \\
	 \therefore T_{10} &= 10 + 3(10)  \\
										&= 40 \\
	 \therefore T_{15} &= 10 + 3(15)  \\
										&= 55 \\
	 \therefore T_{30} &= 10 + 3(30)  \\
										&= 100 
	\end{aligned}\]
----
\(T_n = 10 + 3n\)
----
\(T_{10} = 40\)
----
\(T_{15} = 55\)
----
\(T_{30} = 100\)
----
\(18; 24; 30; 36; \ldots\)
----
\[\begin{aligned}
	d &= T_2 -  T_1  \\
	 &= 24 - 18  \\
	&= 6 \\
	\therefore T_n &= 12 + 6n
	\end{aligned}\]
----
\[\begin{aligned}
	T_n &= 12 + 6n  \\
	 \therefore T_{10} &= 12 + 6(10)  \\
										&=  72 \\
	 \therefore T_{15} &= 12 + 6(15) \\
										&=  102 \\
	 \therefore T_{30} &= 12 + 6(30)  \\
										&=  192
	\end{aligned}\]
----
\(T_n = 12 + 6n\)
----
\(T_{10} = 72\)
----
\(T_{15} = 102\)
----
\(T_{30} = 192\)
----
\(-10; -15; -20; -25; \ldots\)
----
\[\begin{aligned}
	d &= T_2 -  T_1  \\
	 &= -15 -(-10)  \\
	&= -5 \\
	\therefore T_n &= -5 -5n
	\end{aligned}\]
----
\[\begin{aligned}
	T_n &=  -5 -5n \\
	 \therefore T_{10} &= -5 -5(10)  \\
										&=  -55 \\
	 \therefore T_{15} &= -5 -5(15) \\
										&=  -80 \\
	 \therefore T_{30} &= -5 -5(30)  \\
										&=  -155
	\end{aligned}\]
----
\(T_n = -5 -5n\)
----
\(T_{10} = -55\)
----
\(T_{15} = -80\)
----
\(T_{30} = -155\)
----
\[\begin{aligned}
	d &= T_2 -  T_1  \\
	 &= 22-20  \\
	&= 2\\
	\therefore T_n &= 18 + 2n
	\end{aligned}\]
----
\[\begin{aligned}
	T_n &=  18 + 2n\\
	 \therefore T_{9} &= 18 + 2(9)  \\
										&=  18 + 18  \\
										&=  36
	\end{aligned}\]
----
\(T_{9} = 36\)
----
\[\begin{align}
						13 + 31 &= \ldots \\
						24 + 42 &= \ldots \\
						38 + 83 &= \ldots
						\end{align}\]
----
\[\begin{align}
						13 + 31 &= 44 \\
						24 + 42 &= 66 \\
						38 + 83 &= 121
						\end{align}\]
----
\[\begin{align}
						45 + 54 &= 99 \\
						71 + 17 &= 88
						\end{align}\]
----
\(a + 10b\)
----
\(b + 10a\)
----
\[\begin{align}
						\text{Number } 1:  &= a + 10b \\
						\text{Number } 2:  &= b + 10a \\
						\text{Number } 1 + 2:  &= a +b +10a +10b \\
																	  &= 11a +11b \\
																	  &= 11(a + b) 
						\end{align}\]
----
\(n\)
----
\(d\)
----
\(-\)
----
\(T_n = \ldots\)
----
\(n\)
----
\(-\)
----
\(-\)
----
\(-\)
----
\(T_n = \ldots\)
----
\(n\)
----
\(1; 2; 4; 7; 11; \ldots\)
----
\({T}_{n}=a{n}^{2}+bn+c\)
----
\(2a\)
----
\(5; 20; 45; 80; \ldots\)
----
\(6; 11; 18; 27; \ldots\)
----
\(1; 4; 9; 16; \ldots\)
----
\(3; 0; -5; -12; \ldots\)
----
\(1; 3; 7; 13; \ldots\)
----
\(0; -6; -16; -30; \ldots\)
----
\(-1; 2; 9; 20; \ldots\)
----
\(1; -3; -9; -17; \ldots\)
----
\(3a+1; 12a+1; 27a+1; 48a+1  \ldots\)
----
\(2; 10; 24; 44; \ldots\)
----
\(t-2; 4t-1; 9t; 16t+1; \ldots\)
----
\(11; 21; 35; \ldots; 75\)
----
\(T_4 = 53\)
----
\(20; \ldots; 42; 56; 72\)
----
\(T_2 = 30\)
----
\(\ldots; 37; 65; 101\)
----
\(T_1 = 17\)
----
\(3; \ldots; -13; -27; -45\)
----
\(T_2 = -3\)
----
\(24; 35; 48; \ldots; 80\)
----
\(T_4 = 63\)
----
\(\ldots; 11; 26; 47\)
----
\(T_1 = 2\)
----
\(T_n = n^2+3n-1\)
----
\(3;9;17;27\)
----
\(T_n = -n^2-5\)
----
\(-6;-9;-14;-21\)
----
\(T_n = 3n^2-2n\)
----
\(1;8;21;40\)
----
\(T_n = -2n^2+n+1\)
----
\(0;-5;-14;-27\)
----
\(\ldots\)
----
\(T_n = an^2 + bn + c\)
----
\(\ldots\)
----
\(a\)
----
\(b\)
----
\(c\)
----
\(T_n = an^2 + bn + c\)
----
\[\begin{align}
		n=1:  T_1 &= a + b + c \\
		n=2:  T_2 &= 4a + 2b + c \\
		n=3:   T_3 &= 9a + 3b + c
		\end{align}\]
----
\(a\)
----
\(b\)
----
\(c\)
----
\(T_1 = 5\)
----
\(T_2 = 12\)
----
\(T_3 = 23\)
----
\[\begin{align}
		a + b + c  &= 5 \\
		4a + 2b + c  &= 12 \\
		9a + 3b + c  &= 23
		\end{align}\]
----
\[\begin{alignrll}
		T_2 - T_1 &= 4a + 2b + c - (a + b + c) & \\
		12 - 5	  &= 4a + 2b + c - a - b - c   & \\
		7	  &= 3a + b                    & \ldots (1)
		\end{alignrll}\]
----
\[\begin{alignrll}
		T_3 - T_2  &= 9a + 3b + c - (4a + 2b + c) & \\
		23 - 12    &=  9a + 3b + c - 4a - 2b - c  & \\
		11         &=  5a + b                     & \ldots (2)
		\end{alignrll}\]
----
\[\begin{align}
		(2)-(1)  &= 5a + b - (3a + b) \\
		11 - 7 &= 5a + b - 3a - b \\
		4 &= 2a \\
		\therefore a &= 2
		\end{align}\]
----
\[\begin{align}
		\text{Using equation } (1): \quad 3(2) + b &= 7 \\
		\therefore b &= 1 \\
		\text{And using } \quad a + b + c 		 &= 5 \\
		  2 + 1 + c 		 &= 5 \\
		\therefore c &= 1
		\end{align}\]
----
\[T_n = 2n^2 + n + 2\]
----
\[3; 6; 10; 15; 21; \ldots\]
----
\(T_n\)
----
\(T_n\)
----
\(n\)
----
\({3; 4; 5; \ldots}\)
----
\(T_n = an^2 + bn + c\)
----
\(T_n\)
----
\(a\)
----
\(b\)
----
\(c\)
----
\(T_n = an^2 + bn + c\)
----
\[\begin{align}
		n=1:  T_1 &= a + b + c \\
		n=2:  T_2 &= 4a + 2b + c \\
		n=3:   T_3 &= 9a + 3b + c
		\end{align}\]
----
\(a\)
----
\(b\)
----
\(c\)
----
\(T_1 = 3\)
----
\(T_2 = 6\)
----
\(T_3 = 10\)
----
\[\begin{align}
		a + b + c  &= 3 \\
		4a + 2b + c  &= 6 \\
		9a + 3b + c  &= 10
		\end{align}\]
----
\[\begin{alignrll}
		T_2 - T_1  &= 4a + 2b + c - (a + b + c) \\
		6 - 3	   &= 4a + 2b + c - a - b - c  \\
		3	   &= 3a + b  & \ldots (1)
		\end{alignrll}\]
----
\[\begin{alignrll}
		T_3 - T_2  &= 9a + 3b + c - (4a + 2b + c) \\
		10 - 6  &=  9a + 3b + c - 4a - 2b - c  \\
		4  &=  5a + b  & \ldots (2)
		\end{alignrll}\]
----
\[\begin{align}
		(2)-(1)  &= 5a + b - (3a + b) \\
		4 - 3 &= 5a + b - 3a - b \\
		1 &= 2a \\
		\therefore a &= \frac{1}{2}
		\end{align}\]
----
\[\begin{align}
		\text{Using equation }(1): \quad  3\left(\frac{1}{2}\right) + b &= 3 \\
		\therefore b &= \frac{3}{2} \\
		\text{And using } \quad a + b + c 		 &= 3 \\
		\frac{1}{2} + \frac{3}{2} + c 		 &= 3 \\
		\therefore c &= 1
		\end{align}\]
----
\(T_n = \frac{1}{2}n^2 + \frac{3}{2}n + 1\)
----
\(T_n\)
----
\(n\)
----
\(T_n = \frac{1}{2}n^2 + \frac{3}{2}n + 1\)
----
\(n\)
----
\(T_n\)
----
\(n\)
----
\(A\)
----
\(B\)
----
\(C\)
----
\(D\)
----
\(A\)
----
\(AB, AC, AD\)
----
\(B\)
----
\(BC, BD\)
----
\(C\)
----
\(CD\)
----
\(D\)
----
\(3+2+1=6\)
----
\(AB\)
----
\(A\)
----
\(B\)
----
\(BA\)
----
\(AB\)
----
\(A, B, C, D\)
----
\(E\)
----
\(A\)
----
\(AB, AC, AD, AE\)
----
\(B\)
----
\(BC, BD, BE\)
----
\(C\)
----
\(CD, CE\)
----
\(D\)
----
\(DE\)
----
\(E\)
----
\(4+3+2+1=10\)
----
\(A, B, C, D, E\)
----
\(F\)
----
\(A\)
----
\(AB, AC, AD, AE, AF\)
----
\(B\)
----
\(BC, BD, BE, BF\)
----
\(C\)
----
\(CD, CE, CF\)
----
\(D\)
----
\(DE, DF\)
----
\(E\)
----
\(EF\)
----
\(F\)
----
\(5+4+3+2+1=15\)
----
\(T_n = an^2 + bn + c\)
----
\(T_n\)
----
\(a\)
----
\(b\)
----
\(c\)
----
\(T_n = an^2 + bn + c\)
----
\[\begin{align}
		  n=1:  T_1 &= a + b + c \\
		  n=2:  T_2 &= 4a + 2b + c \\
		  n=3:   T_3 &= 9a + 3b + c
		  \end{align}\]
----
\(a\)
----
\(b\)
----
\(c\)
----
\(T_1 = 6\)
----
\(T_2 = 10\)
----
\(T_3 = 15\)
----
\[\begin{align}
		  a + b + c  &= 6 \\
		  4a + 2b + c  &= 10 \\
		  9a + 3b + c  &= 15
		  \end{align}\]
----
\[\begin{alignrll}
		  T_2 - T_1  &= 4a + 2b + c - (a + b + c) \\
		  10 - 6	 &= 4a + 2b + c - a - b - c  \\
		  4	 &= 3a + b  & \ldots (1)
		  \end{alignrll}\]
----
\[\begin{alignrll}
		  T_3 - T_2  &= 9a + 3b + c - (4a + 2b + c) \\
		  15- 10  &=  9a + 3b + c - 4a - 2b - c  \\
		  5  &=  5a + b  & \ldots (2)
		  \end{alignrll}\]
----
\[\begin{align}
		  (2)-(1)  &= 5a + b - (3a + b) \\
		  5-4 &= 5a + b - 3a - b \\
		  1 &= 2a \\
		  \therefore a &= \frac{1}{2}
		  \end{align}\]
----
\[\begin{align}
		  \text{Using equation } (1): \quad  3\left(\frac{1}{2}\right) + b &= 4 \\
		  \therefore b &= \frac{5}{2} \\
									\text{And using } a + b + c 		 &= 6 \\
		  \frac{1}{2} + \frac{5}{2} + c 		 &= 6 \\
		  \therefore c &= 3
		  \end{align}\]
----
\(T_n = \frac{1}{2}n^2 + \frac{5}{2}n + 3\)
----
\(3;\phantom{\rule{0.222222em}{0ex}}6;\phantom{\rule{0.222222em}{0ex}}10;\phantom{\rule{0.222222em}{0ex}}15;\phantom{\rule{0.222222em}{0ex}}21;\phantom{\rule{0.222222em}{0ex}}...\)
----
\(4;\phantom{\rule{0.222222em}{0ex}}9;\phantom{\rule{0.222222em}{0ex}}16;\phantom{\rule{0.222222em}{0ex}}25;\phantom{\rule{0.222222em}{0ex}}36;\phantom{\rule{0.222222em}{0ex}}...\)
----
\(7;\phantom{\rule{0.222222em}{0ex}}17;\phantom{\rule{0.222222em}{0ex}}31;\phantom{\rule{0.222222em}{0ex}}49;\phantom{\rule{0.222222em}{0ex}}71;\phantom{\rule{0.222222em}{0ex}}...\)
----
\(2;\phantom{\rule{0.222222em}{0ex}}10;\phantom{\rule{0.222222em}{0ex}}26;\phantom{\rule{0.222222em}{0ex}}50;\phantom{\rule{0.222222em}{0ex}}82;\phantom{\rule{0.222222em}{0ex}}...\)
----
\(31;\phantom{\rule{0.222222em}{0ex}}30;\phantom{\rule{0.222222em}{0ex}}27;\phantom{\rule{0.222222em}{0ex}}22;\phantom{\rule{0.222222em}{0ex}}15;\phantom{\rule{0.222222em}{0ex}}...\)
----
\(T_n = 5n^2 + 3n + 4\)
----
\[\begin{aligned}
	T_n &= 5n^2 + 3n + 4  \\
	T_1 &= 5(1)^2 + 3(1) + 4  \\
	    &= 12 \\
	T_2 &= 5(2)^2 + 3(2) + 4  \\
	    &= 30 \\
	T_3 &= 5(3)^2 + 3(3) + 4  \\
	    &= 58 \\
	T_4 &= 5(4)^2 + 3(4) + 4  \\
	    &= 96 \\
	T_5 &= 5(5)^2 + 3(5) + 4  \\
	    &= 144 
	\end{aligned}\]
----
\(12;30;58;96;144\)
----
\(T_n = 4n^2 + 5n + 10\)
----
\(T_9\)
----
\[\begin{aligned}
	T_n &=  4n^2 + 5n + 10  \\
	T_9 &=  4(9)^2 + 5(9) + 10   \\
	    &= 379 
	\end{aligned}\]
----
\(T_9 = 379\)
----
\(T_n = 2n^2\)
----
\(n\)
----
\(T_n = 32\)
----
\[\begin{aligned}
	T_n &= 2n^2  \\
	32 &=  2n^2  \\
	 16   &= n^2 \\
 4 &= n
	\end{aligned}\]
----
\(n=4\)
----
\(16; 27; 42; 61; \ldots\)
----
\[\begin{aligned}
	\text{Second difference: } &= 4 \\
	\text{First differences: } &=  11;15;19;23;27 \\
 	\therefore T_5 &=  61+23  \\
 								&= 84   \\
 	\therefore T_6 &= 84 +27   \\
 								&= 111
	\end{aligned}\]
----
\(T_5 = 84; T_6 = 111\)
----
\[\begin{aligned}
	T_n &= an^2 +bn +c  \\
	T_1 &= a +b +c  \\
	T_2 &= 4n^2 +2n +c  \\
	T_3 &= 9n^2 +3n +c 
	\end{aligned}\]
----
\[\begin{aligned}
	\therefore a + b+ c &= 16  \\
	 \therefore c &=  16 - a - b \\
	 4a + 2b + c    &= 27 \\
	 4a + 2b + (16 - a - b)    &= 27 \\
	 3a + b     &= 11 \\
	\text{And } 9a + 3b + c &= 42 \\
 \therefore 9a + 3b + (16 - a - b) &= 42 \\
	8a + 2b = 26 \\
	4a + b =  13 \\
	\therefore b &= 13 -4a
	\end{aligned}\]
----
\[\begin{aligned}
	\therefore 3a + b &= 11 \\
	3a + (13 - 4a) &= 11 \\
	-a  &= -2 \\
	\therefore a  &=  2 \\
	b  &=  13 -4(2) \\
	\therefore b &= 5 \\
  \text{And } c  &= 16 - a - b \\
  \therefore c  &= 16 - 2 -5 \\
			 &= 9 \\
	\therefore T_n  &= 2n^2 + 5n + 9
	\end{aligned}\]
----
\(T_n = 2n^2 + 5n + 9\)
----
\(T_n\)
----
\(d\)
----
\[d = T_n - T_{n-1}\]
----
\[T_n = an^2 + bn + c\]
----
\({T}_{n}={n}^{2}+2n+1\)
----
\(6;9;14;21;30;...\)
----
\[\begin{aligned}
	\text{First differences: } &=  3; 5; 7; 9; \\
	\text{Second difference: } &= 2 
	\end{aligned}\]
----
\(1;7;17;31;49;...\)
----
\[\begin{aligned}
	\text{First differences: } &= 6; 10; 14; 18  \\
	\text{Second difference: } &=  4
	\end{aligned}\]
----
\(8;17;32;53;80;...\)
----
\[\begin{aligned}
	\text{First differences: } &=  9; 15; 21; 27  \\
	\text{Second difference: } &=  6
	\end{aligned}\]
----
\(9;26;51;84;125;...\)
----
\[\begin{aligned}
	\text{First differences: } &=  17; 25; 33; 41  \\
	\text{Second difference: } &=  8
	\end{aligned}\]
----
\(2;20;50;92;146;...\)
----
\[\begin{aligned}
	\text{First differences: } &= 18; 30; 42; 54   \\
	\text{Second difference: } &=   12
	\end{aligned}\]
----
\(5;19;41;71;109;...\)
----
\[\begin{aligned}
	\text{First differences: } &= 14; 22; 30; 38   \\
	\text{Second difference: } &=   8
	\end{aligned}\]
----
\(2;6;10;14;18;...\)
----
\[\text{First difference: } = 4\]
----
\(3;9;15;21;27;...\)
----
\[\text{First difference: } = 6\]
----
\(10;24;44;70;102;...\)
----
\[\begin{aligned}
	\text{First differences: } &=  14; 20; 26; 32 \\
	\text{Second difference: } &=  16
	\end{aligned}\]
----
\(2\frac{1}{2}; 6; 10\frac{1}{2}; 16; 22\frac{1}{2}; \ldots\)
----
\(3p^2; 6p^2; 9p^2; 12p^2; 15p^2; \ldots\)
----
\[\text{First difference: }  = 3p^2\]
----
\(2k; 8k; 18k; 32k; 50k; \ldots\)
----
\[\begin{aligned}
	\text{First differences: } &= 6k; 10k; 14k; 18k   \\
	\text{Second difference: } &= 4k
	\end{aligned}\]
----
\(16; x; 46; \ldots\)
----
\(x\)
----
\[\begin{aligned}
	x - 16 &= 46 - x \\
  	2x &= 62 \\
	\therefore x &= 31 
	\end{aligned}\]
----
\(x = 31\)
----
\(T_n = 2n^2\)
----
\(n\)
----
\(T_n = 242\)
----
\[\begin{aligned}
	2n^2 &= 242 \\
  	n^2 &= 121 \\
	\therefore n &= 11 
	\end{aligned}\]
----
\(n = 11\)
----
\({T}_{n}=3{n}^{2}\)
----
\({T}_{11}\)
----
\[\begin{aligned}
	T_n &= 3n^2 \\
  \therefore T_11	 &= 3(11)^2 \\
	  &= 363 
	\end{aligned}\]
----
\(T_11 = 363\)
----
\(T_n = n^2+4\)
----
\(n\)
----
\(T_n = 85\)
----
\[\begin{aligned}
	n^2 + 4 &= 85 \\
  	n^2 &= 81 \\
	\therefore n &= 9 
	\end{aligned}\]
----
\(n = 9\)
----
\({T}_{n}=4{n}^{2}+3n-1\)
----
\({T}_{5}\)
----
\[\begin{aligned}
	T_n &= 4n^2 + 3n - 1 \\
  \therefore T_5	 &= 4(5)^2 + 3(5) - 1\\
	  &= 100 + 15 - 1 \\
	  &= 114 
	\end{aligned}\]
----
\(T_5 = 114\)
----
\(T_n = \dfrac{3}{2}n^2\)
----
\(n\)
----
\(T_n = 96\)
----
\[\begin{aligned}
	\dfrac{3}{2}n^2  &= 96 \\
  	n^2 &= 96 \times \frac{2}{3} \\
  	 &= 64 \\
	\therefore n &= 8 
	\end{aligned}\]
----
\(n = 8\)
----
\(3; 7; 11; 15; \ldots\)
----
\[\begin{aligned}
	d  &= 7 - 3 \\
  	 &= 4 \\
  	T_5 &= 15 + 4 \\
	      &= 19 \\
  	T_n &= 4n - 1 \\
  	T_{10} &= 4(10) - 1 \\
  	     &= 39
	\end{aligned}\]
----
\(T_5 = 19; T_n = 4n - 1; T_{10} = 39\)
----
\(17; 12; 7; 2;  \ldots\)
----
\[\begin{aligned}
	d  &= 12 - 17  \\
  	 &= -5 \\
  	T_5 &= 2 - 5 \\
	      &= -3 \\
  	T_n &= 22 - 5n \\
  	T_{10} &= 22 - 5(10) \\
  	     &= -28 
	\end{aligned}\]
----
\(T_5 = -3; T_n = 22 - 5n; T_{10} = -28\)
----
\(\frac{1}{2}; 1; 1\frac{1}{2}; 2;   \ldots\)
----
\[\begin{aligned}
	d  &= 1 - \frac{1}{2}  \\
  	 &= \frac{1}{2} \\
  	T_5 &= 2 + \frac{1}{2} \\
	      &= 2 \frac{1}{2} \\
  	T_n &= \frac{1}{2}n \\
  	T_{10} &= \frac{1}{2}(10) \\
  	     &= 5
	\end{aligned}\]
----
\(T_5 = 2\frac{1}{2}; T_n = \frac{1}{2}n; T_{10} = 5\)
----
\(a; a+b; a+2b; a+3b;   \ldots\)
----
\[\begin{aligned}
	d  &= a + b - a  \\
  	 &= b \\
  	T_5 &= a + 3b + b \\
	      &= a + 4b \\
  	T_n &= a - b + bn \\
  	T_{10} &= a - b + b(10) \\
  	     &= a + 9b 
	\end{aligned}\]
----
\(T_5 = a + 4b; T_n = a - b + bn; T_{10} = a + 9b\)
----
\(1; -1; -3; -5;  \ldots\)
----
\[\begin{aligned}
	d  &= -1 - 1  \\
  	 &= -2 \\
  	T_5 &= -5 -2 \\
	      &= -7 \\
  	T_n &= 3 - 2n \\
  	T_{10} &= 3 - 2(10) \\
  	     &= -17 
	\end{aligned}\]
----
\(T_5 = -7; T_n = 3 - 2n; T_{10} = -17\)
----
\({T}_{100}\)
----
\(4;7;12;19;28;...\)
----
\[\begin{aligned}
	T_n &=  an^2 + bn + c \\
	\text{Second difference: } 2a &= 2 \\
														\therefore a &= 1 \\
	3a + b  &=  3 \\
 	b &=  3 - 3(1)  \\
 					\therefore b &= 0   \\
 	a + b + c &= 4   \\
 					\therefore c &= 4 - a - b \\
							&= 4 - 1 \\
\therefore c &= 3 \\
\therefore T_n = n^2 + 3
	\end{aligned}\]
----
\(2;8;14;20;26;...\)
----
\[\begin{aligned}
	d  &=  8 - 2 \\
 	   &=  6  \\
 					\therefore T_n &= 6n - 4 
	\end{aligned}\]
----
\[\begin{aligned}
	T_n &=  6n - 4 \\
 					\therefore T_{100} &= 6(100) - 4  \\
							&= 596
	\end{aligned}\]
----
\(T_n = 6n - 4 ; T_{100} = 596\)
----
\(7;13;23;37;55;...\)
----
\[\begin{aligned}
	T_n &=  an^2 + bn + c \\
	\text{Second difference: } 2a &= 4 \\
														\therefore a &= 2 \\
	3a + b  &=  6 \\
 	b &=  6 - 3(2)  \\
 					\therefore b &= 0   \\
 	a + b + c &= 7   \\
 					\therefore c &= 7 - a - b \\
							&= 7 - 2 \\
\therefore c &= 5 \\
\therefore T_n = 2n^2 + 5
	\end{aligned}\]
----
\(5;14;29;50;77;...\)
----
\[\begin{aligned}
	T_n &=  an^2 + bn + c \\
	\text{Second difference: } 2a &= 6 \\
														\therefore a &= 3 \\
	3a + b  &=  9 \\
 	b &=  9 - 3(3)  \\
 					\therefore b &= 0   \\
 	a + b + c &= 5   \\
 					\therefore c &= 5 - a - b \\
							&= 5 - 3 \\
\therefore c &= 2 \\
\therefore T_n = 3n^2 + 2
	\end{aligned}\]
----
\(T_n = 3n-1\)
----
\(2; 5; 8; 11; 14\)
----
\(d=3\)
----
\(-15; -11; -7; \ldots ; 173\)
----
\[\begin{aligned}
	d &=  -11 - (-15) \\
	  &= 4 \\
	T_n &=  4n - 19
	\end{aligned}\]
----
\(T_n = 4n - 19\)
----
\[\begin{aligned}
	T_n &=  4n - 19 \\
	 173  &= 4n - 19 \\
	192 &=  4n \\
	\therefore 48 &=  n 
	\end{aligned}\]
----
\(n = 48\)
----
\(3; 7; 13; 21; 31; \ldots\)
----
\(T_n = 4n - 1\)
----
\[\begin{aligned}
	\text{First differences: } &= 4; 6; 8; 10   \\
	\text{Second difference: } &= 2
	\end{aligned}\]
----
\(T_n = 4n - 1\)
----
\(T_n = n^2 + n + 1\)
----
\[\begin{aligned}
	T_n &=  an^2 + bn + c \\
	\text{Second difference: } 2a &= 2 \\
														\therefore a &= 1 \\
	3a + b  &=  4 \\
 	b &=  4 - 3(1)  \\
 					\therefore b &= 1   \\
 	a + b + c &= 3   \\
 					\therefore c &= 3 - a - b \\
							&= 3 - 1 - 1 \\
\therefore c &= 1 \\
\therefore T_n &= n^2 + n + 1
	\end{aligned}\]
----
\((n)\)
----
\(n\)
----
\((w)\)
----
\((n)\)
----
\(n\)
----
\((w)\)
----
\(4n - 4\)
----
\((n)\)
----
\(n\)
----
\((h)\)
----
\((n)\)
----
\(n\)
----
\((h)\)
----
\(n+1\)
----
\((n)\)
----
\((c)\)
----
\((n)\)
----
\((c)\)
----
\[\begin{aligned}
	T_n &=  an^2 + bn + c \\
	\text{Second difference: } 2a &= 1 \\
														\therefore a &= \frac{1}{2} \\
	3a + b  &=  3 \\
 	b &=  3 - 3(\frac{1}{2})  \\
 					\therefore b &= \frac{3}{2}   \\
 	a + b + c &= 3   \\
 					\therefore c &= 3 - a - b \\
							&= 3 - \frac{1}{2} - \frac{3}{2} \\
\therefore c &= 1 \\
\therefore T_n &= \frac{1}{2}n^2 + \frac{3}{2}n + 1
	\end{aligned}\]
----
\(T_n = \frac{1}{2}n^2 + \frac{3}{2}n + 1\)
----
\[\begin{aligned}
	T_n &=  \frac{1}{2}n^2 + \frac{3}{2}n + 1 \\  								 \therefore T_{21} &= \frac{1}{2}(21)^{2}+ \frac{3}{2}(21) + 1 \\
							&= \frac{441}{2} + \frac{63}{2} + \frac{2}{2} \\
	 &= 253
	\end{aligned}\]
----
\(T_{21} = 253\)
----
\[\begin{aligned}
	T_n &=  an^2 + bn + c \\
	\text{Second difference: } 2a &= 2 \\
														\therefore a &= 1 \\
	3a + b  &= 2  \\
 	b &=  2 - 3(1)  \\
 					\therefore b &= -1   \\
 	a + b + c &= 0   \\
 					\therefore c &= - a - b \\
							&= -1 - (-1) \\
\therefore c &= 0 \\
\therefore T_n &=  n^2 -n 
	\end{aligned}\]
----
\(T_n =  n^2 - n\)
----
\[\begin{aligned}
	\text{Let } T_2 &=   x \\
	\therefore T_2 - T_1 &=   x - 4 \\
	\text{And } T_3 - T_2 &=   34 - x \\
	\text{Second difference } &=  (T_3 - T_2) - (T_2 - T_1)  \\
	          &=  (34 - x)  - (x - 4)\\
			  \therefore 10 &= 38 - 2x \\
   2x &= 28 \\
	\therefore x &= 14 
	\end{aligned}\]
----
\(4; 14; 34; 64; 104; 154\)
----
\(T_n = an^2 + bn + c\)
----
\(d\)
----
\(D\)
----
\(a = \dfrac{D}{2}\)
----
\[\begin{aligned}
	T_n &=   an^2 + bn +c \\
	T_1 &=   a(1)^2 + b(1) +c \\
	    &=   a + b + c \\
	T_2 &=   a(2)^2 + b(2) +c \\
	    &=   4a + 2b + c\\
	T_3 &=   a(3)^2 + b(3) +c \\
	    &=   9a + 6b + c\\
	\text{First difference } d &=  T_2 - T_1 \\
					\therefore d	 &= (4a + 2b + c) - (a + b + c ) \\
					            	 &= 3a + b \\
					\therefore b &= d - 3a \\
	\text{Second difference } D &=    (T_3 - T_2) - (T_2 - T_1)\\
					\therefore D	 &= (5a + b) - (3a + b) \\
					            	 &= 2a \\
					\therefore a &= \dfrac{D}{2}
	\end{aligned}\]
----
\(b = d - \dfrac{3}{2}D\)
----
\[\begin{aligned}
					  a &= \dfrac{D}{2} \\
					  b &= d - 3a \\
					\therefore b	&= d - 3 \left( \dfrac{D}{2}  \right) \\
		   		            	 &= d -  \dfrac{3}{2} D 
	\end{aligned}\]
----
\(c = T_1 - d + D\)
----
\[\begin{aligned}
	   T_1 &=   a + b + c \\
\therefore c &=   T_1 - a - b \\
					   &= T_1 - \left( \dfrac{D}{2} \right) -  \left( d -  \dfrac{3}{2}D \right) \\
					   &= T_1  -  \dfrac{D}{2}  - d + \dfrac{3}{2}D \\
						 &= T_1 - d + D
	\end{aligned}\]
----
\(T_n = \dfrac{D}{2}n^2 + \left(d - \dfrac{3}{2}D\right)n + \left(T_1 -d + D\right)\)
----
\[\begin{aligned}
	   T_n &=   an^2 + bn +c \\
\therefore T_n  &=   \dfrac{D}{2}n^2 + \left( d - \dfrac{3}{2}D  \right)n + \left( T_1 - d + D \right) 
	\end{aligned}\]
----
\(A(x_1;y_1), B(x_2;y_2)\)
----
\(C(x_2;y_1)\)
----
\[AB^2 = AC^2 + BC^2\]
----
\[AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
----
\({\left({x}_{1}-{x}_{2}\right)}^{2}={\left({x}_{2}-{x}_{1}\right)}^{2}\)
----
\((m)\)
----
\[m_{AB} = \frac{y_2 - y_1}{x_2 - x_1}  \quad \text{ or } \quad m_{AB} = \frac{y_1 - y_2}{x_1 - x_2}\]
----
\(m \ne \dfrac{y_1-y_2}{x_2-x_1}\)
----
\(m = 0\)
----
\(m\)
----
\(m_1 = m_2\)
----
\(m_1 \times m_2 = -1\)
----
\(M(x;y)\)
----
\(A(x_1;y_1)\)
----
\(B(x_2;y_2)\)
----
\[M(x;y) = \left( \frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2} \right)\]
----
\(P(x_1;y_1)\)
----
\(Q(x_2;y_2)\)
----
\(R(x;y)\)
----
\[m_{PR} = m_{QR} = m_{PQ}\]
----
\(m_{PR} = m_{PQ}\)
----
\((x;y)\)
----
\[\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}\]
----
\(P(-5;-4)\)
----
\(Q(0;6)\)
----
\(PQ\)
----
\(T(x;y)\)
----
\(PQ\)
----
\(R(1;-\frac{3}{4})\)
----
\(T(x;y)\)
----
\(PQ\)
----
\(P\)
----
\(\left({x}_{1};{y}_{1}\right)\)
----
\(Q\left({x}_{2};{y}_{2}\right)\)
----
\[x_1 = -5; \qquad y_1 = -4; \qquad x_2 = 0; \qquad y_2 = 6\]
----
\[\begin{align}
		    PQ &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\
		    &= \sqrt{(0 - (-5))^2 + (6-(-4))^2} \\
		    &= \sqrt{25 + 100} \\
		    &= \sqrt{125} \\
		    &= 5\sqrt{5}
		    \end{align}\]
----
\(PQ\)
----
\(5\sqrt{5}\)
----
\[T(x;y) = \left( \frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2} \right)\]
----
\[\begin{align}
		    x &= \frac{x_1 + x_2}{2}   \\
		    &= \frac{-5+0}{2}   \\
  		    &= -\frac{5}{2}   \\
		    y &= \frac{y_1 + y_2}{2}  \\
		    &= \frac{-4+6}{2}\\
		    &= \frac{2}{2} \\
		    &= 1
		    \end{align}\]
----
\(PQ\)
----
\(T(-\frac{5}{2};1)\)
----
\(PQ\)
----
\(RT\)
----
\[m = \frac{y_2 - y_1}{x_2 - x_1}\]
----
\[\begin{align}
		    m_{PQ} &= \frac{6-(-4)}{0-(-5)}   \\
		    &= \frac{10}{5}   \\
  		    &= 2   \\
		    m_{RT} &= \dfrac{-\frac{3}{4} - 1}{1-(-\frac{5}{2})}  \\
		    &= \dfrac{-\frac{7}{4}}{\frac{7}{2}}\\
		    &= -\frac{7}{4} \times \frac{2}{7} \\
		    &= -\frac{1}{2}
		    \end{align}\]
----
\[\begin{align}
		    m_{RT} \times m_{PQ} &= -\frac{1}{2} \times 2  \\
		    &= -1
		    \end{align}\]
----
\(PQ\)
----
\(RT\)
----
\(A\left(-1;0\right)\)
----
\(B\left(0;3\right)\)
----
\(C\left(8;11\right)\)
----
\(D\left(x;y\right)\)
----
\(D\left(x;y\right)\)
----
\(ABCD\)
----
\(AC\)
----
\(BD\)
----
\(AC\)
----
\(D\)
----
\(\left({x}_{1};{y}_{1}\right)\)
----
\(\left({x}_{2};{y}_{2}\right)\)
----
\(AC\)
----
\(M(x;y)\)
----
\[x_1 = -1; \qquad y_1 = 0; \qquad x_2 = 8; \qquad y_2 = 11\]
----
\[M(x;y) = \left( \frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2} \right)\]
----
\(M\)
----
\[\begin{align}
	M(x;y) &=  \left( \frac{-1 + 8}{2}; \frac{0 + 11}{2} \right)   \\
	&= \left( \frac{7}{2}; \frac{11}{2} \right) 
	\end{align}\]
----
\(M\)
----
\(D\)
----
\(M\)
----
\(BD\)
----
\(M\left(\frac{7}{2};\frac{11}{2}\right)\)
----
\(B\left(0;3\right)\)
----
\(D\left(x;y\right)\)
----
\(x\)
----
\(y\)
----
\[\begin{align}
		M &=  \left( \frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2} \right)   \\
		\therefore \left( \frac{7}{2}; \frac{11}{2} \right)  &= \left( \frac{0 + x}{2}; \frac{3 + y}{2} \right) 
		\end{align}\]
----
\[\begin{align}
		\frac{7}{2} &=   \frac{0 + x}{2} \\
		7 &= 0 + x \\
		\therefore x &= 7
		\end{align}\]
----
\[\begin{align}
		\frac{11}{2} &=   \frac{3 + y}{2} \\
		11 &= 3 + y \\
		\therefore y &= 8
		\end{align}\]
----
\(ABCD\)
----
\(D\)
----
\(D\)
----
\(C\)
----
\(A, B\)
----
\(C\)
----
\(BC\)
----
\(AD\)
----
\(B\)
----
\(C\)
----
\(A\)
----
\(D\)
----
\(y = 0 + 8 = 8\)
----
\(B\)
----
\(C\)
----
\(A\)
----
\(D\)
----
\(x = -1 + 8 = 7\)
----
\(AB\)
----
\(DC\)
----
\(A\)
----
\(B\)
----
\(C\)
----
\(D\)
----
\(y = 11 - 3 = 8\)
----
\(A\)
----
\(B\)
----
\(C\)
----
\(D\)
----
\(x = 8 - 1 = 7\)
----
\(D\)
----
\(\left(7;8\right)\)
----
\(P(-3;5)\)
----
\(Q(-1;-5)\)
----
\(T(2x;y-2)\)
----
\(U(3x+1;y-2)\)
----
\(Q(4;1), T(p;3)\)
----
\(QT = \sqrt{8}\)
----
\(p\)
----
\[\begin{aligned}
	QT &= \sqrt{ \left( x_2 - x_1 \right)^2 + \left( y_2 - y_1 \right)^2} \\
	 \sqrt{8}  &= \sqrt{ \left( p-4 \right)^2 + \left( 3-1\right)^2} \\
	\left( \sqrt{8} \right)^2 &=   \left( p-4 \right)^2 + \left( 2 \right)^2 \\
	 8  &=  p^2 - 8p + 16 + 4 \\
	 0  &=  p^2 - 8p + 16 + 4 - 8\\
	  &=  p^2 - 8p + 12 \\
	  &=   \left( p - 6 \right) \left( p - 2 \right)  \\
	 \therefore p  =  6  &\text{ or } p = 2
	\end{aligned}\]
----
\(p = 6\)
----
\(p = 2\)
----
\(AB\)
----
\(A(-5;3)\)
----
\(B(-7;4)\)
----
\(A(3;-2)\)
----
\(B(1;-8)\)
----
\(PQ\)
----
\(P(0;3)\)
----
\(Q(5;5)\)
----
\(5y + 5 = 2x\)
----
\[\begin{aligned}
	5y + 5 &=  2x \\
	  5y &= 2x - 5 \\
	 y  &= \frac{2}{5}x - 1 \\
 \therefore m &= \frac{2}{5}
	\end{aligned}\]
----
\[\begin{aligned}
	m_{PQ} &= \frac{y_2 - y_1}{x_2 - x_1} \\
	  &= \frac{5-3 }{5-0 } \\
	  &= \frac{2}{5}
	\end{aligned}\]
----
\(A(-1;-1)\)
----
\(B(2;5)\)
----
\(C(-1;-\frac{5}{2})\)
----
\(D(x;-4)\)
----
\(AB \perp CD\)
----
\(x\)
----
\(P(x;y)\)
----
\(M(3;5)\)
----
\(N(-1;-1)\)
----
\(A(-3;-4)\)
----
\(B(2;3)\)
----
\(A(-2;4)\)
----
\(B(x;y)\)
----
\(C(1;3)\)
----
\(x\)
----
\(y\)
----
\[\begin{aligned}
	M(x;y) &= \left( \frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2}   \right) \\
	C(1;3) &= \left( \frac{-2 + x}{2}; \frac{4 + y}{2}   \right) \\
   \therefore 1  &=   \frac{-2 + x}{2}  \\
    					2	 &=   -2 + x \\
    			\therefore	4	 &=   x \\
   \text{And } 3  &=  \frac{4 + y}{2}  \\
    					6	 &=   4 + y \\
    			\therefore	2	 &=   y
	\end{aligned}\]
----
\(B(4;2)\)
----
\(ABCD\)
----
\(A(0;3), B(4;3), C(5;-1)\)
----
\(D(1;-1)\)
----
\(AD\)
----
\(BC\)
----
\[\begin{aligned}
	m_{AD} &= \frac{y_2 - y_1}{x_2 - x_1} \\
	  &= \frac{-1-3}{1-0} \\
	  &= \frac{-4}{1} \\
	  &= -4 \\
	 y &= mx + c \\ 
	 y &= -4x + c \\ 
	\text{Subst. } (0;3) \quad 3 &= -4(0) + c \\ 
	 \therefore c &= 3 \\ 
	 y &= -4x + 3  
	\end{aligned}\]
----
\[\begin{aligned}
	m_{BC} &= \frac{y_2 - y_1}{x_2 - x_1} \\
	  &= \frac{-1-3}{5-4} \\
	  &= \frac{-4}{1} \\
	  &= -4 \\
	 y &= mx + c \\ 
	 y &= -4x + c \\ 
	\text{Subst. } (4;3) \quad 3 &= -4(4) + c \\ 
	 \therefore c &= 19 \\ 
	 y &= -4x + 19
	\end{aligned}\]
----
\(y  = -4x + 3\)
----
\(y = -4x + 19\)
----
\(AD \parallel BC\)
----
\[\begin{aligned}
	m_{AD} &= -4 \\
	m_{BC} &= -4 \\
 \therefore m_{AD} &= m_{BC} \\
 \therefore AD &\parallel BC 
	\end{aligned}\]
----
\(AD\)
----
\(BC\)
----
\(BD\)
----
\[\begin{aligned}
	m_{BD} &= \frac{y_2 - y_1}{x_2 - x_1} \\
	  &= \frac{-1 -3}{1-4} \\
	  &= \frac{-4}{-3} \\
	  &= \frac{4}{3}\\
	 y &= mx + c \\ 
	 y &= \frac{4}{3}x + c \\ 
	\text{Subst. } (4;3) \quad 3 &= \frac{4}{3}(4) + c \\ 
	  &= 3 - \frac{16}{3} \\ 
	 \therefore c &=  - \frac{7}{3} \\ 
	 y &= \frac{4}{3}x - \frac{7}{3}
	\end{aligned}\]
----
\(y = \frac{4}{3}x - \frac{7}{3}\)
----
\(ABCD\)
----
\(MPQN\)
----
\(M(-5;3)\)
----
\(P(-1;5)\)
----
\(Q(4;5)\)
----
\(N(x;y)\)
----
\(MPQN\)
----
\(PQ = MN\)
----
\(PQ \parallel MN\)
----
\[\begin{aligned}
	m_{PQ} &= \frac{y_2 - y_1}{x_2 - x_1} \\
	  &= \frac{5 - 5}{4 + 1} \\
	  &= 0 \\
	 \therefore m_{MN} &=  0\\
	 \therefore y &=  3 \\
	 PQ &= \sqrt{ \left( x_2 - x_1 \right)^2 + \left( y_2 - y_1 \right)^2} \\
	    &= \sqrt{ \left( 4+1 \right)^2 + \left( 5 - 5 \right)^2} \\
	    &= \sqrt{ \left( 5 \right)^2} \\
	    &=  5 \\
	 \text{Therefore } x &= -5 + 5 \\ 
	  					\therefore x	&= 0 \\ 
	 \therefore N(x;y) &=  (0;3)
	\end{aligned}\]
----
\(N(0;3)\)
----
\(PQRS\)
----
\(P(-3;1)\)
----
\(Q(1;3)\)
----
\(R(6;1)\)
----
\(S(2;-1)\)
----
\(PQ\)
----
\(SR\)
----
\[\begin{aligned}
	 PQ &= \sqrt{ \left( x_2 - x_1 \right)^2 + \left( y_2 - y_1 \right)^2} \\
	     &= \sqrt{ \left( 1 + 3 \right)^2 + \left(3 - 1 \right)^2} \\
	     &= \sqrt{ \left( 4 \right)^2 + \left( 2 \right)^2} \\
	     &= \sqrt{ 16 + 4} \\
	     &= \sqrt{20} \\
	 SR &= \sqrt{ \left( x_2 - x_1 \right)^2 + \left( y_2 - y_1 \right)^2} \\
	     &= \sqrt{ \left( 2- 6 \right)^2 + \left( -1-1 \right)^2} \\
	     &= \sqrt{ \left( -4 \right)^2 + \left( -2 \right)^2} \\
	     &= \sqrt{ 16 + 4} \\
	     &= \sqrt{20}
	\end{aligned}\]
----
\(PQ = \sqrt{20}\)
----
\(SR = \sqrt{20}\)
----
\(PR\)
----
\(PR\)
----
\(M(x;y)\)
----
\[\begin{aligned}
	M(x;y) &= \left( \frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2}   \right) \\
					&= \left( \frac{-3 + 6}{2}; \frac{1 + 1 }{2}   \right) \\
					&= \left( \frac{3}{2}; 1  \right) 
	\end{aligned}\]
----
\(M(\frac{3}{2}; 1)\)
----
\(PQ \parallel SR\)
----
\[\begin{aligned}
	m_{PQ} &= \frac{y_2 - y_1}{x_2 - x_1} \\
	  &= \frac{3 - 1}{1 + 3} \\
	  &= \frac{2}{4} \\
	  &= \frac{1}{2} \\
	m_{SR} &= \frac{y_2 - y_1}{x_2 - x_1} \\
	  &= \frac{-1-1}{2-6} \\
	  &= \frac{-2}{-4} \\
	  &= \frac{1}{2} \\
	 \text{Therefore } m_{PQ} &= m_{SR}\\ 
	  						\therefore PQ &\parallel SR 
	\end{aligned}\]
----
\(PS\)
----
\(SR\)
----
\[\begin{aligned}
	m_{PS} &= \frac{y_2 - y_1}{x_2 - x_1} \\
	  &= \frac{-1-1}{2+3} \\
	  &= \frac{-2}{5} \\
	 y &= mx + c \\ 
	 y &= -\frac{2}{5}x + c \\ 
	\text{Subst. } (-3;1) \quad 1 &= -\frac{2}{5}(-3) + c \\ 
    1 &= \frac{6}{5} + c \\ 
	  \therefore -\frac{1}{5} &= c \\ 
	 y &= -\frac{2}{5}x - \frac{1}{5}
	\end{aligned}\]
----
\[\begin{aligned}
	m_{SR} &= \frac{1}{2} \\
	 y &= mx + c \\ 
	 y &= \frac{1}{2}x + c \\ 
	\text{Subst. } (6;1) \quad 1 &= \frac{1}{2}(6) + c \\ 
    1 &= 3 + c \\ 
	  \therefore -2 &= c \\ 
	 y &= \frac{1}{2}x - 2
	\end{aligned}\]
----
\(PS: y = -\frac{2}{5}x - \frac{1}{5}\)
----
\(SR: y = \frac{1}{2}x - 2\)
----
\(PS \perp SR\)
----
\(m_{PS} \times m_{SR} \ne -1\)
----
\(PQRS\)
----
\(\dfrac{y-y_1}{x-x_1} = \dfrac{y_2-y_1}{x_2-x_1}\)
----
\(y-y_1 = m(x-x_1)\)
----
\(y = mx + c\)
----
\(\left({x}_{1};{y}_{1}\right)\)
----
\(\left({x}_{2};{y}_{2}\right)\)
----
\[\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\]
----
\(P\left(-1;-5\right)\)
----
\(Q\left(5;4\right)\)
----
\(P\)
----
\((x_1;y_1)\)
----
\(Q(x_2;y_2)\)
----
\[x_1 = -1; \qquad y_1 = -5; \qquad x_2 = 5; \qquad y_2 = 4\]
----
\[\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\]
----
\(y\)
----
\[\begin{align}
							\frac{y-(-5)}{x-(-1)} &= \frac{4-(-5)}{5-(-1)}   \\
							\frac{y+5}{x+1} &= \frac{9}{6}   \\
							y+5 &= \frac{3}{2}(x+1)   \\
							y+5 &= \frac{3}{2}x + \frac{3}{2} \\
							y &= \frac{3}{2}x - \frac{7}{2} 
							\end{align}\]
----
\[y = \frac{3}{2}x - 3\frac{1}{2}\]
----
\((3;7)\)
----
\((-6;1)\)
----
\[\begin{aligned}
	\frac{y-y_1}{x-x_1} &=  \frac{y_2-y_1}{x_2-x_1} \\
	\frac{y-7}{x-3} &=  \frac{1-7}{-6-3} \\
  \frac{y-7}{x-3} &=  \frac{-6}{-9} \\
  \frac{y-7}{x-3} &=  \frac{2}{3} \\
  y - 7 &=  \frac{2}{3} \left( x - 3 \right) \\
  y &=  \frac{2}{3}x - 2 + 7  \\
  \therefore	 y &= \frac{2}{3}x + 5
	\end{aligned}\]
----
\(y = \frac{2}{3}x + 5\)
----
\((1;-\frac{11}{4})\)
----
\((\frac{2}{3}; -\frac{7}{4})\)
----
\[\begin{aligned}
	\frac{y-y_1}{x-x_1} &=  \frac{y_2-y_1}{x_2-x_1} \\
	\frac{y + \frac{11}{4}}{x-1} &=  \frac{-\frac{7}{4} + \frac{11}{4}}{\frac{2}{3} - 1} \\
	\frac{y + \frac{11}{4}}{x-1} &=  \frac{1}{-\frac{1}{3}} \\
	\frac{y + \frac{11}{4}}{x-1} &=  -3 \\
	 y + \frac{11}{4} &=  -3 (x - 1) \\
	 y &=  -3x + 3 - \frac{11}{4} \\
	 \therefore y &=  -3x + \frac{1}{4} 
	\end{aligned}\]
----
\(y =  -3x + \frac{1}{4}\)
----
\((-2;1)\)
----
\((3;6)\)
----
\[\begin{aligned}
	\frac{y-y_1}{x-x_1} &=  \frac{y_2-y_1}{x_2-x_1} \\
	\frac{y-1}{x+2} &=  \frac{6-1}{3+2} \\
  \frac{y-1}{x+2} &=  \frac{5}{5} \\
  y - 1 &=  x + 2 \\
  \therefore  y &=   x + 3  
	\end{aligned}\]
----
\(y = x + 3\)
----
\((2;3)\)
----
\((3;5)\)
----
\[\begin{aligned}
	\frac{y-y_1}{x-x_1} &=  \frac{y_2-y_1}{x_2-x_1} \\
	\frac{y-3}{x-2} &=  \frac{5-3}{3-2} \\
  \frac{y-3}{x-2} &=  2 \\
  y - 3 &=  2(x - 2) \\
    y  &=  2x - 4 + 3\\
   \therefore y &=  2x - 1
	\end{aligned}\]
----
\(y = 2x - 1\)
----
\((1;-5)\)
----
\((-7;-5)\)
----
\[\begin{aligned}
	\frac{y-y_1}{x-x_1} &=  \frac{y_2-y_1}{x_2-x_1} \\
	\frac{y+5}{x-1} &=  \frac{-5+5}{-7-1} \\
  \frac{y+5}{x-1} &=  0 \\
  y + 5 &=  0\\
  \therefore y &=  -5
	\end{aligned}\]
----
\(y = -5\)
----
\((-4;0)\)
----
\((1;\frac{15}{4})\)
----
\[\begin{aligned}
	\frac{y-y_1}{x-x_1} &=  \frac{y_2-y_1}{x_2-x_1} \\
	\frac{y-0}{x+4} &=  \frac{\frac{15}{4}-0}{1+4} \\
	\frac{y}{x+4} &=  \frac{\frac{15}{4}}{5} \\
	\frac{y}{x+4} &=   \frac{15}{20} \\
  y &=   \frac{3}{4} \left( x+4 \right) \\
  \therefore y &=   \frac{3}{4}x + 3
	\end{aligned}\]
----
\(y = \frac{3}{4}x + 3\)
----
\((s;t)\)
----
\((t;s)\)
----
\[\begin{aligned}
	\frac{y-y_1}{x-x_1} &=  \frac{y_2-y_1}{x_2-x_1} \\
	\frac{y-t}{x-s} &=  \frac{s - t}{t - s}\\
	\frac{y-t}{x-s} &=  \frac{-(t - s)}{t - s}\\
	\frac{y-t}{x-s} &=   -1\\
	y - t &=   -(x - s)\\
	y &=   -x + s + t 
	\end{aligned}\]
----
\(y =    -x + (s + t)\)
----
\((-2;-8)\)
----
\((1;7)\)
----
\[\begin{aligned}
	\frac{y-y_1}{x-x_1} &=  \frac{y_2-y_1}{x_2-x_1} \\
	\frac{y+8}{x+2} &=  \frac{7+8}{1+2} \\
	\frac{y+8}{x+2} &=  \frac{15}{3} \\
	\frac{y+8}{x+2} &=  5 \\
	y + 8 &=   5(x + 2)\\
	y + 8 &=   5x + 10\\
	y &=  5x + 2  
	\end{aligned}\]
----
\(y = 5x + 2\)
----
\((2p;q)\)
----
\((0;-q)\)
----
\[\begin{aligned}
	\frac{y-y_1}{x-x_1} &=  \frac{y_2-y_1}{x_2-x_1} \\
	\frac{y-q}{x-2p} &=  \frac{-q-q}{0 - 2p} \\
	\frac{y-q}{x-2p} &=  \frac{-2q}{-2p} \\
	\frac{y-q}{x-2p} &=  \frac{q}{p} \\
	y - q &=  \frac{q}{p} \left( x - 2p \right) \\
	y - q &=  \frac{q}{p}x - 2q \\
	\therefore y  &=  \frac{q}{p}x - q 
	\end{aligned}\]
----
\(y =  \frac{q}{p}x - q\)
----
\[\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\]
----
\(m = \dfrac{y_2-y_1}{x_2-x_1}\)
----
\[\frac{y-y_1}{x-x_1} = m\]
----
\((x-x_1)\)
----
\[y-y_1 = m(x-x_1)\]
----
\(m = -\frac{1}{3}\)
----
\((-1;1)\)
----
\(m < 0\)
----
\(x\)
----
\[y - y_1= m(x - x_1)\]
----
\[y - y_1= -\frac{1}{3}(x - x_1)\]
----
\[\begin{align}
							y - 1 &= -\frac{1}{3}(x - (-1))   \\
							y - 1 &= -\frac{1}{3}(x + 1)   \\
							y &= -\frac{1}{3}x -  \frac{1}{3} + 1\\
							  &= -\frac{1}{3}x + \frac{2}{3}
							\end{align}\]
----
\(y = -\frac{1}{3}x +\frac{2}{3}\)
----
\(\left(-3;2\right)\)
----
\(\left(5;8\right)\)
----
\[x_1 = -3; \qquad y_1 = 2; \qquad x_2 = 5; \qquad y_2 = 8\]
----
\[\begin{align}
	m &= \frac{y_2 - y_1}{x_2 - x_1}  \\
	&= \frac{8 - 2}{5 - (- 3)}  \\
	&= \frac{6}{8}  \\
	&= \frac{3}{4}  
	\end{align}\]
----
\[y - y_1= m(x - x_1)\]
----
\[y - y_1= \frac{3}{4}(x - x_1)\]
----
\[\begin{align}
							y - y_1 &= \frac{3}{4}(x - x_1)   \\
							y - 2 &= \frac{3}{4}(x - (-3))   \\
							y - 2 &= \frac{3}{4}(x + 3)   \\
							y &= \frac{3}{4}x +  \frac{9}{4} + 2\\
							  &= \frac{3}{4}x + \frac{17}{4}
							\end{align}\]
----
\(y = \frac{3}{4}x +4\frac{1}{4}\)
----
\((-1;\frac{10}{3})\)
----
\(m=\frac{2}{3}\)
----
\[\begin{aligned}
	y - y_1 &= m(x - x_1)\\
	y - \frac{10}{3} &= \frac{2}{3}(x + 1)\\
	y - \frac{10}{3} &= \frac{2}{3}x + \frac{2}{3} \\
	y &= \frac{2}{3}x + \frac{2}{3} + \frac{10}{3} \\
 \therefore y  &= \frac{2}{3}x + 4  
	\end{aligned}\]
----
\(y = \frac{2}{3}x + 4\)
----
\(m= -1\)
----
\((-2;0)\)
----
\[\begin{aligned}
	y - y_1 &= m(x - x_1) \\
	y - 0 &= -(x + 2) \\
	\therefore y &= -x - 2  
	\end{aligned}\]
----
\(y = -x - 2\)
----
\((3;-1)\)
----
\(m=-\frac{1}{3}\)
----
\[\begin{aligned}
	y - y_1 &= m(x - x_1) \\
	y + 1 &= -\frac{1}{3}(x - 3) \\
	y + 1 &= -\frac{1}{3}x + 1 \\
	\therefore y &= -\frac{1}{3}x  
	\end{aligned}\]
----
\(y = -\frac{1}{3}x\)
----
\(x\)
----
\((0;11)\)
----
\[\begin{aligned}
	y - y_1 &= m(x - x_1) \\
	y - 11 &= 0(x - 0) \\
	\therefore y &= 11
	\end{aligned}\]
----
\(y = 11\)
----
\((1;5)\)
----
\(m=-2\)
----
\[\begin{aligned}
	y - y_1 &= m(x - x_1) \\
	y - 5 &= -2(x - 1) \\
	y - 5 &=-2x + 2 \\
	\therefore y &= -2x + 7
	\end{aligned}\]
----
\(y = -2x + 7\)
----
\(x\)
----
\((-\frac{3}{2};0)\)
----
\(x = -\frac{3}{2}\)
----
\((10;-7)\)
----
\[\begin{aligned}
	y - y_1 &= m(x - x_1) \\
	y + 7 &= -\frac{4}{5}(x - 10) \\
	y + 7 &= -\frac{4}{5}x + 8 \\
	\therefore y &= -\frac{4}{5}x + 1
	\end{aligned}\]
----
\(y = -\frac{4}{5}x + 1\)
----
\((4;0)\)
----
\(x = 4\)
----
\(m= 3a\)
----
\((-2;-6a+b)\)
----
\[\begin{aligned}
	y - y_1 &= m(x - x_1) \\
	y - (-6a + b) &= 3a(x + 2) \\
	y + 6a - b &= 3ax + 6a \\
	\therefore y  &= 3ax + b
	\end{aligned}\]
----
\(y = 3ax + b\)
----
\[y - y_1 = m(x - x_1)\]
----
\(y\)
----
\[\begin{align}
	y - y_1 &= mx - mx_1  \\
	y &= mx - mx_1 + y_1  \\
	y &= mx + (y_1 - mx_1) 
	\end{align}\]
----
\(c\)
----
\(c={y}_{1}-m{x}_{1}\)
----
\[y = mx + c\]
----
\(x=0\)
----
\[\begin{align}
  y &= m(0) + c \\
  &= c
  \end{align}\]
----
\(c\)
----
\(y\)
----
\(m=-2\)
----
\((-1;7)\)
----
\(m < 0\)
----
\(x\)
----
\[y =mx + c\]
----
\[y = -2x + c\]
----
\(c\)
----
\[\begin{align}
	y &= -2x + c \\
	7 &= -2(-1) + c   \\
	7 - 2 &= c \\
	\therefore c &= 5
	\end{align}\]
----
\(y\)
----
\((0;5)\)
----
\(y = -2x + 5\)
----
\(m\)
----
\(c\)
----
\((-2;-7)\)
----
\((3;8)\)
----
\[y =mx + c\]
----
\[\begin{array}{rll}
          -7 & = m(-2) + c & \\
          -7 & = -2m + c & \ldots (1) \\
          \\
          8 & = m(3) + c & \\
          8 & = 3m + c & \ldots (2)
          \end{array}\]
----
\(c\)
----
\(c\)
----
\[\begin{align}
  -7 &= -2m + c \\
  -( 8 &= 3m + c )  \\
  -15 &= -5m  \\
  \therefore 3 &= m  
  \end{align}\]
----
\(m=3\)
----
\(c\)
----
\[\begin{align}
    -7 &= -2m + c \\
    -7 &= -2(3) + c \\
    \therefore c &= -1
    \end{align}\]
----
\[\begin{align}
    8 &= 3m + c \\
    8 &= 3(3) + c \\
    \therefore c &= -1
    \end{align}\]
----
\(y = 3x - 1\)
----
\((\frac{1}{2};4)\)
----
\(m=2\)
----
\[\begin{aligned}
	y  &= mx + c \\
		y  &= 2x + c \\
		4  &= 2\left( \frac{1}{2} \right) + c \\
	4 &= 1 + c \\
	\therefore c  &= 3 \\
	\therefore y  &= 2x + 3 
	\end{aligned}\]
----
\(y = 2x + 3\)
----
\((\frac{1}{2};-2)\)
----
\((2;4)\)
----
\[\begin{aligned}
	y  &= mx + c \\
		-2  &= m\left( \frac{1}{2} \right) + c \\
		\text{Multiply by } 2: \quad -4  &= m  + 2c \ldots (1)\\
		4  &= 2m + c   \\
		\text{Multiply by } 2: \quad 8  &= 4m  + 2c \ldots (2)\\
	(2) - (1): \quad 8 + 4 &= 4m - m  \\
	12 &= 3m  \\
	\therefore 4  &= m  \\
	\therefore c  &= 4 - 2(4) \\
 								 &= -4 \\
	\therefore y  &= 4x - 4 
	\end{aligned}\]
----
\(y = 4x - 4\)
----
\((2;-3)\)
----
\((-1;0)\)
----
\[\begin{aligned}
	y  &= mx + c \\
		-3  &= 2m  + c \ldots (1) \\
  	0  &= -m + c  \ldots (2) \\
	(1) - (2): \quad -3 &= 2m + m  \\
	-3 &= 3m  \\
	\therefore -1  &= m  \\
	\therefore c  &= -1 \\
	\therefore y  &= -x - 1 
	\end{aligned}\]
----
\(y = -x - 1\)
----
\((2;-\frac{6}{7})\)
----
\(m=-\frac{3}{7}\)
----
\[\begin{aligned}
	y  &= mx + c \\
	y  &= -\frac{3}{7}x + c \\
	-\frac{6}{7}  &= -\frac{3}{7}(2) + c \\
	-\frac{6}{7} + \frac{6}{7}  &= + c \\
		\therefore c &= 0 \\
	\therefore y  &= -\frac{3}{7}x  
	\end{aligned}\]
----
\(y = -\frac{3}{7}x\)
----
\(y\)
----
\(y=-\frac{1}{5}\)
----
\(m=\frac{1}{2}\)
----
\[\begin{aligned}
	y  &= mx + c \\
 y  &= mx - \frac{1}{5} \\
	\therefore y  &= \frac{1}{2}x -\frac{1}{5}  
	\end{aligned}\]
----
\(y = \frac{1}{2}x -\frac{1}{5}\)
----
\[\begin{aligned}
	y  &= mx + c \\
 	2  &= 2m + c \ldots (1) \\
 	-4  &= -m + c \ldots (2) \\
	(1)-(2) \quad 2 + 4 &= 2m + m \\
	  6 &= 3m \\
	\therefore m &= 2 \\
	\therefore c &= -4 +2 \\
  &= -2\\
	\therefore y  &= 2x -2 
	\end{aligned}\]
----
\(y = 2x -2\)
----
\(y = -\frac{3}{2}\)
----
\[\begin{aligned}
	c  &= 4 \\
	y &= mx + c \\
	y &= mx + 4 \\
	\text{Subst. } (-2;-2) \quad -2 &= -2m + 4 \\
	 -2m &= -6 \\
	 m &= 3 \\
	\therefore y  &= 3x + 4
	\end{aligned}\]
----
\(y = 3x + 4\)
----
\[\begin{aligned}
	c  &= 0 \\
	y &= mx + c \\
	y &= mx + 0\\
	\text{Subst. } (-2;10) \quad 10 &= -2m  \\
	 -2m &= 10 \\
	 \therefore m &= -5 \\
	\therefore y  &= -5x
	\end{aligned}\]
----
\(y = -5x\)
----
\(\theta\)
----
\(x\)
----
\(\theta\)
----
\(y\)
----
\(x\)
----
\[m=\frac{\Delta y}{\Delta x}\]
----
\[\tan \theta = \frac{\text{opposite side}}{\text{adjacent side}}\]
----
\(x\)
----
\(x\)
----
\(\Delta x = 0\)
----
\(\tan \theta\)
----
\(\tan \theta\)
----
\(y\)
----
\(\Delta y = 0\)
----
\(\tan \theta\)
----
\(\tan \theta\)
----
\(m < 0\)
----
\(\tan \theta < 0\)
----
\(x\)
----
\(m = \dfrac{3}{4}\)
----
\(2y - x = 6\)
----
\((-4;-1)\)
----
\((2;5)\)
----
\(y=4\)
----
\(x = 3y + \frac{1}{2}\)
----
\((2;5)\)
----
\((\frac{2}{3};1)\)
----
\((2;1)\)
----
\((-3;-9)\)
----
\[x_1 = 2; \qquad y_1 = 1; \qquad x_2 = -3; \qquad y_2 = -9\]
----
\[\begin{align}
							m &= \frac{y_2 - y_1}{x_2 - x_1} \\
							 &=  \frac{-9-1}{-3-2} \\
							 &= \frac{-10}{-5} \\
							\therefore m &= 2
							\end{align}\]
----
\((3;1)\)
----
\[y - y_1 = m(x - x_1)\]
----
\(m = -1\)
----
\[y - y_1 = -(x - x_1)\]
----
\((3;1)\)
----
\[\begin{align}
							 y - 1 & = -(x - 3) \\
							y & = -x + 3 +1 \\
							&= -x + 4
							\end{align}\]
----
\(y = -x + 4\)
----
\(M(-1;1\frac{3}{4})\)
----
\(N(4;3)\)
----
\(y = - \frac{3}{2}x + 4\)
----
\(M(-1;1\frac{3}{4})\)
----
\(N(4;3)\)
----
\(y = - \frac{3}{2}x + 4\)
----
\(\alpha\)
----
\(\beta\)
----
\(\theta\)
----
\(\alpha\)
----
\(\theta\)
----
\(\beta\)
----
\(\beta\)
----
\(y = -\frac{3}{2}x + 4\)
----
\(m < 0\)
----
\(\beta\)
----
\(\beta\)
----
\(M\)
----
\(N\)
----
\[\begin{align}
		  m & = \frac{y_2 - y_1}{x_2 - x_1} \\
		  & = \dfrac{3 - \frac{7}{4}}{4-(-1)}  \\
		  & = \dfrac{\frac{5}{4}}{5}  \\
		  &= \frac{1}{4}
		  \end{align}\]
----
\(m = \dfrac{4}{5}\)
----
\(x + y + 1 = 0\)
----
\((1;1)\)
----
\((-2;7)\)
----
\(3 - 2y = 9x\)
----
\((-1;-6)\)
----
\((-\frac{1}{2};-\frac{11}{2})\)
----
\(5 = 10y - 15x\)
----
\(A(-2;\frac{1}{5})\)
----
\(B(0;1)\)
----
\(C(1;0)\)
----
\(D(-2;6)\)
----
\(AB\)
----
\(\beta\)
----
\(CD\)
----
\(\alpha\)
----
\(\theta\)
----
\(y + x = 3\)
----
\(x = y + \frac{1}{2}\)
----
\(y + x = 3\)
----
\(\alpha\)
----
\(x = y + \frac{1}{2}\)
----
\(\beta\)
----
\(\theta\)
----
\(y = 2x\)
----
\((-1;\frac{7}{3})\)
----
\((0;2)\)
----
\(y = 2x\)
----
\(\beta\)
----
\(\alpha\)
----
\(\theta\)
----
\(P(-1;0)\)
----
\(Q(1;4)\)
----
\(R(1;2)\)
----
\(S(2;4)\)
----
\(\alpha\)
----
\(\beta\)
----
\(PQ\)
----
\(RS\)
----
\(\alpha\)
----
\(\beta\)
----
\(\alpha\)
----
\(\beta\)
----
\(PQ \parallel RS\)
----
\(\tan \alpha\)
----
\(\tan \beta\)
----
\(\ldots \ldots\)
----
\(\ldots \ldots\)
----
\(PQ\)
----
\(RS\)
----
\(m_{PQ}\)
----
\(m_{RS}\)
----
\(\ldots \ldots\)
----
\(\ldots \ldots\)
----
\(\left({x}_{1};{y}_{1}\right)\)
----
\(y = m_1x + c_1\)
----
\(y = m_2x + c_2\)
----
\[m_1 = m_2\]
----
\(x\)
----
\[y=mx+c\]
----
\(m_2\)
----
\(\left({x}_{1};{y}_{1}\right)\)
----
\[y - y_1 = m(x - x_1)\]
----
\((-1;1)\)
----
\(y - 2x + 1 = 0\)
----
\(m\)
----
\[y = 2x - 1\]
----
\(m_1 = m_2 = 2\)
----
\[y - y_1 = m(x - x_1)\]
----
\(m = 2\)
----
\[y - y_1 = 2(x - x_1)\]
----
\((-1;1)\)
----
\[\begin{align}
							 y - 1 & = 2(x - (-1)) \\
							y - 1& = 2x + 2\\
							y & = 2x + 2 + 1\\
							&= 2x + 3
							\end{align}\]
----
\(y = 2x + 3\)
----
\(AB\)
----
\(A(0;3)\)
----
\(CD\)
----
\(C(2;-3)\)
----
\(AB\)
----
\(AB \parallel CD\)
----
\[y - y_1 = m(x - x_1)\]
----
\[y - y_1 = -\frac{1}{2}(x - x_1)\]
----
\((2;-3)\)
----
\[\begin{align}
							 y - (-3) & = -\frac{1}{2}(x - 2) \\
							y  + 3 & = -\frac{1}{2}x + 1\\
							y & = -\frac{1}{2}x - 2
							\end{align}\]
----
\(y = -\frac{1}{2}x - 2\)
----
\(y + 2 x = 1\)
----
\(-2x + 3 = y\)
----
\[\begin{aligned}
y + 2 x &= 1 \\ 
	y &= -2x + 1   \\
	\therefore m_1 &= -2 \\
-2x + 3 &= y \\ 
	\therefore m_2 &= -2 \\
 \therefore m_1 &= m_2
	\end{aligned}\]
----
\(\therefore\)
----
\(\frac{y}{3} + x + 5 = 0\)
----
\(2y + 6x = 1\)
----
\[\begin{aligned}
\frac{y}{3} + x + 5 &= 0\\ 
	\frac{y}{3} &= -x - 5   \\
	y &= -3x - 15   \\
	\therefore m_1 &= -3 \\
2y + 6x &= 1 \\ 
2y  &= -6x + 1 \\ 
y &= -3x + \frac{1}{3} \\ 
	\therefore m_2 &= -3 \\
 \therefore m_1 &= m_2
	\end{aligned}\]
----
\(\therefore\)
----
\(y = 2x - 7\)
----
\((1;-2)\)
----
\((\frac{1}{2};-1)\)
----
\[\begin{aligned}
y  &=  2x - 7 \\ 
	\therefore m_1 &= 2 \\
 m_2 &= \frac{y_2 - y_1}{x_2 - x_1} \\
   &= \frac{-1 + 2}{\frac{1}{2} - 1} \\
   &= \frac{1}{\frac{1}{2}} \\
	\therefore m_2 &= 2 \\
 \therefore m_1 &= m_2
	\end{aligned}\]
----
\(\therefore\)
----
\(y + 1 = x\)
----
\(x + y = 3\)
----
\[\begin{aligned}
y + 1 &= x\\ 
	y &= x - 1  \\
	\therefore m_1 &= 1\\
x + y &= 3 \\ 
y  &= -x + 3\\ 
	\therefore m_2 &= - \\
 \therefore m_1 & \ne m_2
	\end{aligned}\]
----
\(\therefore\)
----
\((-2;-1)\)
----
\((-4;-3)\)
----
\(-y + x - 4 = 0\)
----
\[\begin{aligned}
 - y + x - 4 &=  0 \\ 
   y  &=  x - 4 \\ 
	\therefore m_1 &= 1 \\
 m_2 &= \frac{y_2 - y_1}{x_2 - x_1} \\
   &= \frac{-3 + 1}{-4 + 2} \\
   &= \frac{-2}{-2} \\
	\therefore m_2 &= 1 \\
 \therefore m_1 &= m_2
	\end{aligned}\]
----
\(\therefore\)
----
\(y - 1 = \frac{1}{3}x\)
----
\((-2;4)\)
----
\((1;5)\)
----
\[\begin{aligned}
 y - 1 &=  \frac{1}{3}x \\ 
   y  &= \frac{1}{3}x + 1 \\ 
	\therefore m_1 &= \frac{1}{3} \\
 m_2 &= \frac{y_2 - y_1}{x_2 - x_1} \\
   &= \frac{5 - 4}{1 + 2} \\
   &= \frac{1}{3} \\
 \therefore m_1 &= m_2
	\end{aligned}\]
----
\(\therefore\)
----
\((1;-5)\)
----
\(y + 2x - 1 = 0\)
----
\[\begin{aligned}
 y + 2x - 1  &= 0 \\ 
   y  &= -2x + 1 \\ 
	\therefore m &= -2  \\
 y - y_1&= m(x - x_1) \\
 y + 5 &= -2(x - 1) \\
 y &= -2x + 2  - 5  \\
 \therefore  y &= -2x - 3  
	\end{aligned}\]
----
\(y = -2x - 3\)
----
\((-2;-6)\)
----
\(2y + 1 = 6x\)
----
\[\begin{aligned}
 2y + 1  &= 6x \\ 
   2y  &= 6x - 1 \\ 
   y  &= 3x - \frac{1}{2} \\ 
	\therefore m &= 3  \\
 y - y_1&= m(x - x_1) \\
 y + 6 &= 3(x + 2) \\
 y &= 3x + 6 - 6 \\
 \therefore  y &= 3x 
	\end{aligned}\]
----
\(y = 3x\)
----
\((-2;-2)\)
----
\(y = \frac{3}{2}x + 1\)
----
\((-2;\frac{2}{5})\)
----
\(y = -\frac{7}{10}x -1\)
----
\(A(-2;-3)\)
----
\(B(2;5)\)
----
\(C(-1;\frac{1}{2})\)
----
\(D(4;-2)\)
----
\(\alpha\)
----
\(\beta\)
----
\(AB\)
----
\(CD\)
----
\(\theta\)
----
\(AB\)
----
\(CD\)
----
\(AB\)
----
\(CD\)
----
\(\theta\)
----
\(AB \perp CD\)
----
\(AB\)
----
\(CD\)
----
\(\tan \alpha \times \tan \beta\)
----
\(m_{AB} \times m_{CD}\)
----
\(\ldots \ldots\)
----
\(\ldots \ldots\)
----
\(\ldots \ldots\)
----
\(\ldots \ldots\)
----
\(\ldots \ldots\)
----
\(m_1 \times m_2 = -1\)
----
\(A(4;3)\)
----
\(A\hat{O}X = \theta\)
----
\(B\)
----
\((-3;4)\)
----
\(OA\)
----
\[\begin{align}
	m_{OA} &= \frac{y_2 - y_1}{x_2 - x_1}  \\
	&= \frac{3 - 0}{4 - 0} \\
	&= \frac{3}{4}
	\end{align}\]
----
\(OB\)
----
\[\begin{align}
	m_{OB} &= \frac{y_2 - y_1}{x_2 - x_1}  \\
	&= \frac{4 - 0}{-3 - 0} \\
	&= \frac{4}{-3}
	\end{align}\]
----
\(OB \perp OA\)
----
\[\begin{align}
	m_{OA} \times m_{OB} &= \frac{3}{4} \times \frac{4}{-3}  \\
	&= -1 
	\end{align}\]
----
\[m_{OA} = -\frac{1}{m_{OB}}\]
----
\(A(x;y)\)
----
\(A\hat{O}X = \theta\)
----
\(B(-y;x)\)
----
\[\begin{align}
		  m_{OA} &= \frac{y}{x}  \\
		  m_{OB} &= -\frac{x}{y}  \\
		  \therefore m_{OA} \times m_{OB}&= \frac{y}{x} \times -\frac{x}{y}  \\
		  &= -1
		  \end{align}\]
----
\(\left({x}_{1};{y}_{1}\right)\)
----
\(y = m_1x + c_1\)
----
\(y = m_2x + c_2\)
----
\[m_1 \times m_2 = -1\]
----
\(x\)
----
\(y = mx + c\)
----
\[m_1  = -\frac{1}{m_2}\]
----
\(m_1\)
----
\(\left({x}_{1};{y}_{1}\right)\)
----
\(y-y_1 = m(x-x_1)\)
----
\(T(2;2)\)
----
\(3y + 2x - 6 = 0\)
----
\(m_1\)
----
\(m_2\)
----
\(m_2\)
----
\[\begin{align}
		  3y + 2x - 6 &= 0 \\
		  3y & = -2x + 6 \\
		  y & = -\frac{2}{3}x + 2\\
		  \therefore m_2 & = -\frac{2}{3}
		  \end{align}\]
----
\(m_1 \times m_2 = -1\)
----
\(m_1 = \frac{3}{2}\)
----
\[y - y_1 = m(x - x_1)\]
----
\(m_1 = \frac{3}{2}\)
----
\[y - y_1 = \frac{3}{2}(x - x_1)\]
----
\(T(2;2)\)
----
\[\begin{align}
		  y - 2 & = \frac{3}{2}(x - 2) \\
		  y - 2& = \frac{3}{2}x - 3\\
		  y & = \frac{3}{2}x - 1
		  \end{align}\]
----
\(y = \frac{3}{2}x - 1\)
----
\((2;\frac{1}{3})\)
----
\(m_1\)
----
\(m_2\)
----
\[\begin{align}
		   m_{1} \times m_{2} &= -1 \\
		   \therefore m_{1} &= - \frac{1}{3}
		 \end{align}\]
----
\[y - y_1 = m(x - x_1)\]
----
\(m_{1} = -\frac{1}{3}\)
----
\[y - y_1 = -\frac{1}{3}(x - x_1)\]
----
\((2;\frac{1}{3})\)
----
\[\begin{align}
		 y - \left(\frac{1}{3}\right) & = -\frac{1}{3}(x - 2) \\
		 y  -\frac{1}{3} & = -\frac{1}{3}x + \frac{2}{3}\\
		 y & = -\frac{1}{3}x + 1
		 \end{align}\]
----
\(y = -\frac{1}{3}x + 1\)
----
\(y - 1 = 4x\)
----
\(4y + x + 2 = 0\)
----
\[\begin{aligned}
y - 1 &= 4x \\ 
	y &= 4x + 1   \\
	\therefore m_1 &= 4\\
 4y + x + 2 &= 0 \\ 
 4y &= - x - 2 \\ 
 y &= - \frac{1}{4}x - \frac{1}{2} \\ 
	\therefore m_2 &= -\frac{1}{4} \\
 \therefore m_1 \times m_2&= -\frac{1}{4} \times 4 \\
                            &= -1 
	\end{aligned}\]
----
\(\therefore\)
----
\(10x = 5y - 1\)
----
\(5y - x - 10 = 0\)
----
\[\begin{aligned}
10x  &= 5y - 1 \\ 
	10x + 1 &= 5y   \\
	2x + \frac{1}{5} &= y   \\
	\therefore m_1 &= 2 \\
 5y - x - 10 &= 0 \\ 
 5y &= - x + 10 \\ 
 y &= - \frac{1}{5}x + 2 \\ 
	\therefore m_2 &= -\frac{1}{5} \\
 \therefore m_1 \times m_2&= -\frac{1}{5} \times 2 \\
                        &\ne -1  
	\end{aligned}\]
----
\(\therefore\)
----
\(x = y - 5\)
----
\((-1;\frac{5}{4})\)
----
\((3;-\frac{11}{4})\)
----
\[\begin{aligned}
x  &= y - 5 \\ 
	x + 5 &= y   \\
	\therefore m_1 &= 1 \\
 m_2 &= \frac{y_2 - y_1}{x_2 - x_1} \\ 
     &= \frac{-\frac{11}{4} -\frac{5}{4}}{3 + 1} \\
     &= \frac{-4}{4} \\
	\therefore m_2 &= -1 \\
 \therefore m_1 \times m_2&= -1 \times 1 \\
                        &= -1  
	\end{aligned}\]
----
\(\therefore\)
----
\(y = 2\)
----
\(x = 1\)
----
\(\frac{y}{3} = x\)
----
\(3y + x = 9\)
----
\[\begin{aligned}
\frac{y}{3}  &= x \\ 
	y &= 3x   \\
	\therefore m_1 &= 3 \\
3y + x &= 9 \\ 
	3y &= -x + 9  \\
	y &= -\frac{1}{3}x + 3  \\
	\therefore m_2 &= -\frac{1}{3} \\
 \therefore m_1 \times m_2&= -\frac{1}{3} \times 3 \\
                        &= -1  
	\end{aligned}\]
----
\(\therefore\)
----
\(1 - 2x = y\)
----
\((2;-1)\)
----
\((-1;5)\)
----
\[\begin{aligned}
1 -2x  &= y \\ 
	\therefore m_1 &= -2 \\
 m_2 &= \frac{y_2 - y_1}{x_2 - x_1} \\ 
     &= \frac{5 + 1}{-1-2} \\
     &= \frac{6}{-3} \\
	\therefore m_2 &= -2 \\
 \therefore m_1 \times m_2&= -2 \times -2 \\
                        &\ne -1  
	\end{aligned}\]
----
\(\therefore\)
----
\(y = x + 2\)
----
\(2y + 1 = 2x\)
----
\[\begin{aligned}
 y  &= x + 2 \\ 
	\therefore m_1 &= 1 \\
2y + 1 &= 2x \\ 
	2y &= 2x - 1 \\
	y &= x - \frac{1}{2} \\
	\therefore m_2 &= 1 \\
 \therefore m_1 \times m_2&= 1 \times 1 \\
                        &\ne -1  
	\end{aligned}\]
----
\(\therefore\)
----
\((-2;-4)\)
----
\(y + 2x = 1\)
----
\[\begin{aligned}
 y  &= -2x + 1 \\ 
	\therefore m_1 &= -2 \\
 \text{For } \perp:  m_1 \times m_2 &= -1 \\
-2 \times m_2 &= -1 \\
	\therefore m_2 &= \frac{1}{2} \\
y &= mx + c \\ 
y &= \frac{1}{2}x + c \\
\text{Subst. } (-2;-4): \quad - 4 &= \frac{1}{2}(-2) + c \\
	-4 &= -1 + c \\
	\therefore c &= -3 \\
	\therefore y &= \frac{1}{2}x - 3  
	\end{aligned}\]
----
\(y = \frac{1}{2}x - 3\)
----
\((2;-7)\)
----
\(5y - x = 0\)
----
\[\begin{aligned}
 5y - x   &= 0 \\
 5y &= x \\
    y &= \frac{1}{5}x \\
	\therefore m_1 &= \frac{1}{5} \\
 \text{For } \perp:  m_1 \times m_2 &= -1 \\
\frac{1}{5} \times m_2 &= -1 \\
	\therefore m_2 &= -5 \\
y &= mx + c \\ 
y &= -5x + c \\
\text{Subst. } (2;7): \quad -7 &= -5(2) + c \\
	-7 &= -10 + c \\
	\therefore c &= 3 \\
	\therefore y &= -5x + 3  
	\end{aligned}\]
----
\(y = -5x + 3\)
----
\((3;-1)\)
----
\(y = -x + 2\)
----
\((-2;\frac{2}{5})\)
----
\(y = \frac{4}{3}\)
----
\(x = -2\)
----
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
----
\(m = \dfrac{y_2 - y_1}{x_2 - x_1}\)
----
\(M(x;y) = \left( \dfrac{x_1 + x_2}{2}; \dfrac{y_1 + y_2}{2} \right)\)
----
\(m_1 = m_2\)
----
\(m_1 \times m_2 = -1\)
----
\(ax + by + c = 0\)
----
\(\dfrac{y - y_1}{x - x_1} = \dfrac{y_2 - y_1}{x_2 - x_1}\)
----
\(y - y_1 = m(x - x_1)\)
----
\(y = mx + c\)
----
\(\theta\)
----
\(x\)
----
\(m = \tan \theta\)
----
\(\left(-1;3\right)\)
----
\(\left(1;4\right)\)
----
\[\begin{aligned}
m &= \frac{y_2 - y_1}{x_2 - x_1}\\
  &= \frac{4-3}{1 +1}\\
 \therefore m &= \frac{1}{2} \\
    y &= mx + c \\
	\therefore  y &= \frac{1}{2}x + c \\
\text{Subst. } (1;4): \quad 4 &= \frac{1}{2}(1) + c \\
	\therefore c &= 3\frac{1}{2} \\
	\therefore y &= \frac{1}{2}x + \frac{7}{2} 
	\end{aligned}\]
----
\(y=\frac{1}{2}x + \frac{7}{2}\)
----
\(\left(7;-3\right)\)
----
\(\left(0;4\right)\)
----
\[\begin{aligned}
m &= \frac{y_2 - y_1}{x_2 - x_1}\\
  &= \frac{4+3}{0-7} \\
  &= \frac{7}{-7} \\
 \therefore m &= -1 \\
    y &= mx + c \\
	\therefore  y &= -x + c \\
\text{Subst. } (0;4): \quad 4 &= -1(0) + c \\
	\therefore c &= 4 \\
	\therefore y &= -x + 4
	\end{aligned}\]
----
\(y=-x + 4\)
----
\(y=\frac{1}{2}x+3\)
----
\(\left(-2;3\right)\)
----
\[\begin{aligned}
 \therefore m &= \frac{1}{2} \\
    y &= mx + c \\
	\therefore  y &= \frac{1}{2}x + c \\
\text{Subst. } (-2;3): \quad 3 &= \frac{1}{2}(-2) + c \\
	\therefore c &= 4 \\
	\therefore y &= \frac{1}{2}x + 4
	\end{aligned}\]
----
\(y=\frac{1}{2}x + 4\)
----
\(y=-\frac{1}{2}x+3\)
----
\(\left(-1;2 \right)\)
----
\[\begin{aligned}
 \therefore m &= 2 \\
    y &= mx + c \\
	\therefore  y &= 2x + c \\
\text{Subst. } (-1;2): \quad 2 &= 2(-1) + c \\
	\therefore c &= 4 \\
	\therefore y &= 2x + 4
	\end{aligned}\]
----
\(y=2x + 4\)
----
\(3y+x=6\)
----
\[\begin{aligned}
 3y + x &= 6 \\
 y &= -\frac{1}{3}x + 2 \\
 \therefore m &= 3 \\
    y &= mx + c \\
	\therefore  y &= 3x + c \\
\text{Subst. } (0;0): \quad 0 &= 3(0) + c \\
	\therefore c &= 0 \\
	\therefore y &= 3x
	\end{aligned}\]
----
\(y=3x\)
----
\(y=2x-3\)
----
\(y=\frac{1}{3}x-7\)
----
\(4y=3x+8\)
----
\(y=-\frac{2}{3}x+3\)
----
\(3y+x-3=0\)
----
\(P(2;3)\)
----
\(Q(-4;0)\)
----
\(R(5;-3)\)
----
\(\triangle PQR\)
----
\(PR\)
----
\(x\)
----
\(S\)
----
\(PR\)
----
\[\begin{aligned}
m &= \frac{y_2 - y_1}{x_2 - x_1}\\
  &= \frac{-3-3}{5-2} \\
  &= \frac{-6}{3} \\
 \therefore m &= -2 \\
    y &= mx + c \\
	\therefore  y &= -2x + c \\
\text{Subst. } (2;3): \quad 3 &= -2(2) + c \\
	\therefore c &= 7 \\
	\therefore y &= -2x + 7
	\end{aligned}\]
----
\(y=-2x + 7\)
----
\(S\)
----
\[\begin{aligned}
	y &= -2x + 7 \\
	0 &= -2x + 7 \\
	\therefore x &= \frac{7}{2} \\
	\therefore  &S\left( \frac{7}{2}; 0 \right)
	\end{aligned}\]
----
\(\left( \frac{7}{2}; 0 \right)\)
----
\(PR\)
----
\(PQ\)
----
\[\begin{aligned}
m &= \frac{y_2 - y_1}{x_2 - x_1}\\
  &= \frac{3-0}{2+4} \\
  &= \frac{3}{6} \\
 \therefore m &= \frac{1}{2}
	\end{aligned}\]
----
\(m=\frac{1}{2}\)
----
\(Q\hat{P}R\)
----
\(PQ\)
----
\[\begin{aligned}
m_{PQ} &= \frac{1}{2}\\
  \therefore m_{\perp}&= -2 \\
  y &= -2x + c \\
  c &= 0 \\
 \therefore y &= -2x
	\end{aligned}\]
----
\(y=-2x\)
----
\(M\)
----
\(QR\)
----
\[\begin{aligned}
M(x;y) &= \left( \frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2} \right) \\
  &= \left( \frac{-4+5}{2}; \frac{0-3}{2} \right) \\
  &= \left( \frac{1}{2}; -\frac{3}{2} \right)
	\end{aligned}\]
----
\(\left( \frac{1}{2}; -\frac{3}{2} \right)\)
----
\(PR\)
----
\(M\)
----
\[\begin{aligned}
m &= -2 \\
  y &= mx + c \\
  y &= -2x + c \\
\text{Subst. } \left( \frac{1}{2}; -\frac{3}{2} \right): \quad -\frac{3}{2} &= -2\left( \frac{1}{2} \right) + c \\
  c &= -\frac{1}{2} \\
  y &= - 2x -\frac{1}{2}
	\end{aligned}\]
----
\(y=-2x - \frac{1}{2}\)
----
\(A(-3;5)\)
----
\(B(-7;-4)\)
----
\(C(2;0)\)
----
\(D\)
----
\(ABCD\)
----
\[\begin{aligned}
m_{BC} &= \frac{-4-0}{-7-2} \\
    &= \frac{4}{9} \\
  \therefore m_{AD} &= \frac{4}{9} \\
\therefore \text{ from } A(-3;5): &\quad 9 \text{ units to the right and } 4 \text{ units up}
	\end{aligned}\]
----
\(D \left( 6;9 \right)\)
----
\(ABCD\)
----
\[\begin{aligned}
AB &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\
    &= \sqrt{(-7-(-3))^2 + (-4-5)^2} \\
    &= \sqrt{(-4)^2 + (-9)^2} \\
    &= \sqrt{16 + 81} \\
    &= \sqrt{97} \\
AD &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\
    &= \sqrt{(-3-6)^2 + (5-9)^2} \\
    &= \sqrt{(-9)^2 + (-4)^2} \\
    &= \sqrt{81 + 16} \\
    &= \sqrt{97} \\
    \therefore AB &= AD \\
\therefore \text{ parallelogram } ABCD &\text{ is a rhombus (adj. sides equal ) }
	\end{aligned}\]
----
\(y = -x -3\)
----
\(y = 3\)
----
\(y = \frac{5}{2}x + \frac{1}{2}\)
----
\(N\)
----
\[\begin{aligned}
\frac{5}{2}x + \frac{1}{2} &= -x - 3 \\
\frac{5}{2}x + x &= - 3 - \frac{1}{2} \\
  \frac{7}{2}x &= -\frac{7}{2} \\
 \therefore x &= -1 \\
  \therefore y &= -(-1) - 3 \\
    &= -2 \\
  \therefore  &N\left( -1;-2 \right)
	\end{aligned}\]
----
\(\left( -1;-2 \right)\)
----
\(P\)
----
\[\begin{aligned}
y &= \frac{5}{2}x + \frac{1}{2} \\
y &= \frac{5}{2}(3) + \frac{1}{2} \\
    &= 8 \\
  \therefore  &P\left( 8;3 \right)
	\end{aligned}\]
----
\(\left( 8;3 \right)\)
----
\(MN\)
----
\(x = -1\)
----
\(MN\)
----
\(M\hat{N}P\)
----
\(NP\)
----
\(M\)
----
\[\begin{aligned}
    m  &= \frac{5}{2} \\
 y &=mx +c \\
 y &=\frac{5}{2}x +c \\
\text{Subst. } M(-1;3): \quad 3 &= \frac{5}{2}(-1) + c \\
	\therefore c &= 3 + \frac{5}{2} \\
	  &=  \frac{11}{2} \\
 \therefore y &=\frac{5}{2}x + \frac{11}{2}  
	\end{aligned}\]
----
\(y = \frac{5}{2}x + \frac{11}{2}\)
----
\(A(-2;3)\)
----
\(B(2;4)\)
----
\(C(3;0)\)
----
\(\triangle ABC\)
----
\[\begin{aligned}
BC &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\
    &= \sqrt{(3-2)^2 + (0-4)^2} \\
    &= \sqrt{(1)^2 + (4)^2} \\
    &= \sqrt{1 + 16} \\
    &= \sqrt{17} \\
AB &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\
    &= \sqrt{(2+2)^2 + (4-3)^2} \\
    &= \sqrt{(4)^2 + (1)^2} \\
    &= \sqrt{16 + 1} \\
    &= \sqrt{17} \\
    \therefore BC &= AB \\
\therefore \triangle ABC &\text{ is isosceles triangle } \\
AC &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\
    &= \sqrt{(3+2)^2 + (0-3)^2} \\
    &= \sqrt{(5)^2 + (-3)^2} \\
    &= \sqrt{25+9} \\
    &= \sqrt{34} \\
 \text{And } BC^2 + AB^2 &= 17 + 17 \\
   &= 34 \\
   &= AC^2 \\
 \therefore \triangle ABC &\text{ is right-angled triangle}  
	\end{aligned}\]
----
\(AB\)
----
\[\begin{aligned}
    m  &= \frac{y_2-y_1}{x_2-x_1} \\
     &= \frac{4-3}{2+2} \\
      &= \frac{1}{4} \\
 y &=mx +c \\
 y &=\frac{1}{4}x +c \\
\text{Subst. } M(-2;3): \quad 3 &= \frac{1}{4}(-2) + c \\
	\therefore c &= \frac{7}{2} \\
 \therefore y &=\frac{1}{4}x + \frac{7}{2}  
	\end{aligned}\]
----
\(y = \frac{1}{4}x + \frac{7}{2}\)
----
\(D\)
----
\(ABCD\)
----
\(D(-1;-1)\)
----
\(E\)
----
\(BC\)
----
\[\begin{aligned}
E(x;y) &= \left( \frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2} \right) \\
  &= \left( \frac{2+3}{2}; \frac{4+0}{2} \right) \\
  &= \left( \frac{5}{2}; 2 \right)
	\end{aligned}\]
----
\(E \left( \frac{5}{2}; 2 \right)\)
----
\(S(2;5)\)
----
\(T(-3;-4)\)
----
\(V(4;-2)\)
----
\(ST\)
----
\[\begin{aligned}
    m  &= \frac{y_2-y_1}{x_2-x_1} \\
     &= \frac{-4-5}{-3-2} \\
      &= \frac{-9}{-6} \\
      &= \frac{3}{2} \\
 y &=mx +c \\
 y &=\frac{3}{2}x +c \\
\text{Subst. } M(2;5): \quad 5 &= \frac{3}{2}(2) + c \\
	\therefore c &= 2\\
 \therefore y &=\frac{3}{2}x + 2  
	\end{aligned}\]
----
\(y = \frac{3}{2}x + 2\)
----
\(T\hat{S}V\)
----
\(FGH\)
----
\(F(-1;3)\)
----
\(G(2;1)\)
----
\(H(4;4)\)
----
\(\triangle FGH\)
----
\(\triangle FGH\)
----
\[\begin{aligned}
FG &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\
    &= \sqrt{(-1-2)^2 + (3-1)^2} \\
    &= \sqrt{(-3)^2 + (2)^2} \\
    &= \sqrt{9 + 4} \\
    &= \sqrt{13} \\
GH &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\
    &= \sqrt{(4-2)^2 + (4-1)^2} \\
    &= \sqrt{(2)^2 + (3)^2} \\
    &= \sqrt{4 + 9} \\
    &= \sqrt{13} \\
    \therefore FG &= GH \\
\therefore \triangle FGH &\text{ is isosceles triangle } 
	\end{aligned}\]
----
\(PQ\)
----
\(FH\)
----
\[\begin{aligned}
Q(x;y) &= \left( \frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2} \right) \\
  &= \left( \frac{4-1}{2}; \frac{4+3}{2} \right) \\
  &= \left( \frac{3}{2}; \frac{7}{2} \right) \\
    m  &= \frac{y_2-y_1}{x_2-x_1} \\
     &= \frac{4-3}{4+1} \\
      &= \frac{1}{5} \\
 \therefore m_{\perp} &= -5  \\
 y &=mx +c \\
 y &=-5x +c \\
\text{Subst. } Q\left( \frac{3}{2}; \frac{7}{2} \right): \quad \frac{7}{2} &= -5 \left( \frac{3}{2} \right) + c \\
	\therefore c &= 11 \\
 \therefore y &= -5x + 11  
	\end{aligned}\]
----
\(y = -5x + 11\)
----
\(G\)
----
\(PQ\)
----
\[\begin{aligned}
 y &=-5x + 11 \\
 G(2;1) \therefore \text{subst. } x=2: \quad y &= -5 \left( 2 \right) + 11 \\
 &= -10 + 11 \\
 &= 1 \\
 \text{Yes, } G & \text{ lies on PQ }
	\end{aligned}\]
----
\(GH\)
----
\(F\)
----
\[\begin{aligned}
    m_{GH}  &= \frac{y_2-y_1}{x_2-x_1} \\
     &= \frac{4-1}{4-2} \\
      &= \frac{3}{2} \\
 y &=mx +c \\
 y &=\frac{3}{2}x +c \\
\text{Subst. } F\left( -1;3 \right): \quad 3 &=   \frac{3}{2}(-1) + c \\
	\therefore c &= \frac{9}{2} \\
 \therefore y &= \frac{3}{2}x + \frac{9}{2}  
	\end{aligned}\]
----
\(y = \frac{3}{2}x + \frac{9}{2}\)
----
\(A(-1;5)\)
----
\(B(5;-3)\)
----
\(C(0;-6)\)
----
\(M\)
----
\(AB\)
----
\(N\)
----
\(AC\)
----
\(M\)
----
\(N\)
----
\((2;1)\)
----
\((-\frac{1}{2};-\frac{1}{2})\)
----
\[\begin{aligned}
M(x;y) &= \left( \frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2} \right) \\
  &= \left( \frac{-1+5}{2}; \frac{5-3}{2} \right) \\
  &= \left( \frac{4}{2}; \frac{2}{2} \right) \\
  &= \left( 2; 1 \right) \\
N(x;y) &= \left( \frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2} \right) \\
  &= \left( \frac{-1+0}{2}; \frac{5-6}{2} \right) \\
  &= \left( -\frac{1}{2}; -\frac{1}{2} \right)  
	\end{aligned}\]
----
\(NM \parallel CB\)
----
\(NM = \frac{1}{2}CB\)
----
\[\begin{aligned}
    m_{NM}  &= \frac{y_2-y_1}{x_2-x_1} \\
     &= \dfrac{1+\frac{1}{2}}{2+\frac{1}{2}} \\
      &= \dfrac{\frac{3}{2}}{\frac{5}{2}} \\
      &= \frac{3}{5} \\
    m_{CB}  &= \frac{y_2-y_1}{x_2-x_1} \\
     &= \frac{-3+6}{5-0} \\
      &= \frac{3}{5} \\
      m_{NM} &= m_{CB} \\
 \therefore NM &\parallel CB \\
NM &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\
    &= \sqrt{(2+\frac{1}{2})^2 + (1+\frac{1}{2})^2} \\
    &= \sqrt{(\frac{5}{2})^2 + (\frac{3}{2})^2} \\
    &= \sqrt{\frac{25}{4} + \frac{9}{4}} \\
    &= \sqrt{\frac{34}{4}} \\
    &= \frac{1}{2}\sqrt{34} \\
CB &= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\
    &= \sqrt{(5-0)^2 + (-3+6)^2} \\
    &= \sqrt{(5)^2 + (3)^2} \\
    &= \sqrt{25 + 9} \\
    &= \sqrt{34} \\
    \therefore NM &= \frac{1}{2}CB  
	\end{aligned}\]
----
\(y = a x^2 + q\)
----
\(y=a{x}^{2}+q\)
----
\(a\)
----
\(q\)
----
\(a\)
----
\(q\)
----
\(f(x) = ax^2 + q\)
----
\(q\)
----
\(q>0\)
----
\(f(x)\)
----
\(q\)
----
\(f(x)\)
----
\(x\)
----
\(q<0\)
----
\(f(x)\)
----
\(q\)
----
\(f(x)\)
----
\(x\)
----
\(q\)
----
\(y\)
----
\(a\)
----
\(a>0\)
----
\(f(x)\)
----
\((0;q)\)
----
\(a\)
----
\(a\)
----
\(f(x)\)
----
\(a<0\)
----
\(f(x)\)
----
\((0;q)\)
----
\(a\)
----
\(a\)
----
\(f(x)\)
----
\(a<0\)
----
\(a>0\)
----
\(q>0\)
----
\(q=0\)
----
\(q<0\)
----
\(y_1 = x^2\)
----
\(y_2 = \frac{1}{2}x^2\)
----
\(y_3 = -x^2 - 1\)
----
\(y_4 = -2x^2 + 4\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(q\)
----
\(q = 0\)
----
\(q\)
----
\(y_{\text{int}} = 0\)
----
\(a\)
----
\(a = 1\)
----
\(a\)
----
\((0;0)\)
----
\(x = 0\)
----
\(y\)
----
\(\{x: x \in \mathbb{R} \}\)
----
\(\{y: y \geq 0\}\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(q\)
----
\(q\)
----
\(y_{\text{int}} = 0\)
----
\(y_{\text{int}} = 0\)
----
\(y_{\text{int}} = -1\)
----
\(y_{\text{int}} = 4\)
----
\(a\)
----
\(a = 1\)
----
\(a = \frac{1}{2}\)
----
\(a = -1\)
----
\(a = -2\)
----
\(a\)
----
\((0;0)\)
----
\((0;0)\)
----
\((0;-1)\)
----
\((0;4)\)
----
\(x = 0\)
----
\(y\)
----
\(x = 0\)
----
\(y\)
----
\(x = 0\)
----
\(y\)
----
\(x = 0\)
----
\(y\)
----
\(\{x: x \in \mathbb{R} \}\)
----
\(\{x: x \in \mathbb{R} \}\)
----
\(\{x: x \in \mathbb{R} \}\)
----
\(\{x: x \in \mathbb{R} \}\)
----
\(\{y: y \geq 0\}\)
----
\(\{y: y \geq 0\}\)
----
\(\{y: y \leq -1\}\)
----
\(\{y: y \leq 4\}\)
----
\(y=a{\left(x+p\right)}^{2}+q\)
----
\(y=a{\left(x+p\right)}^{2}+q\)
----
\(p\)
----
\(a\)
----
\(p\)
----
\(q\)
----
\(y_1 = x^2\)
----
\(y_2 = (x - 2)^2\)
----
\(y_3 = (x - 1)^2\)
----
\(y_4 = (x + 1)^2\)
----
\(y_5 = (x + 2)^2\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(y_5\)
----
\(x\)
----
\(y\)
----
\(p\)
----
\(y_1 = x^2 + 2\)
----
\(y_2 = (x - 2)^2 - 1\)
----
\(y_3 = (x - 1)^2 + 1\)
----
\(y_4 = (x + 1)^2 + 1\)
----
\(y_5 = (x + 2)^2 - 1\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(y_5\)
----
\(x\)
----
\(y\)
----
\(q\)
----
\(y_1 = (x - 2)^2 + 1\)
----
\(y_2 = 2(x - 2)^2 + 1\)
----
\(y_3 = -\frac{1}{2}(x - 2)^2 + 1\)
----
\(a\)
----
\(y_2\)
----
\(y_1\)
----
\(y_2\)
----
\(y_1\)
----
\(y_2\)
----
\(a\)
----
\(y_3\)
----
\(y_3\)
----
\(y_1\)
----
\(y_3\)
----
\(y_1\)
----
\(y_3\)
----
\(y = a(x + p)^2 + q\)
----
\(p\)
----
\(p>0\)
----
\(p\)
----
\(p<0\)
----
\(p\)
----
\(p\)
----
\(y\)
----
\(\left(p>0\right)\)
----
\(y\)
----
\(\left(p<0\right)\)
----
\(x = -p\)
----
\(q\)
----
\(q\)
----
\(x\)
----
\(\left(q>0\right)\)
----
\(x\)
----
\(\left(q<0\right)\)
----
\(a\)
----
\(a<0\)
----
\(a>0\)
----
\(a = 0\)
----
\(y = q\)
----
\(p>0\)
----
\(p<0\)
----
\(a<0\)
----
\(a>0\)
----
\(a<0\)
----
\(a>0\)
----
\(q>0\)
----
\(q<0\)
----
\(f(x) = y = a(x + p)^2 + q\)
----
\(\left\{x:x\in ℝ\right\}\)
----
\(x\)
----
\(f(x)\)
----
\(f(x)\)
----
\(a\)
----
\(a>0\)
----
\[\begin{array}{r@{\;}c@{\;}l@{\quad}l}
	        (x + p)^2      & \geq & 0 & (\text{perfect square is always positive}) \\
	        \therefore a(x + p)^2     & \geq & 0 & (a \text{ is positive}) \\
	        \therefore a(x + p)^2 + q & \geq & q & \\
	        \therefore f(x)           & \geq & q &
	      \end{array}\]
----
\(\{ y: y \geq q, y \in \mathbb{R} \}\)
----
\(a > 0\)
----
\(a < 0\)
----
\(\{ y: y \leq q, y \in \mathbb{R} \}\)
----
\(g(x) = -2(x - 1)^2 + 3\)
----
\(\{x: x \in \mathbb{R} \}\)
----
\(x\)
----
\(g(x)\)
----
\(g(x)\)
----
\[\begin{align*}
  (x - 1)^2& \geq 0 \\
  -2(x - 1)^2& \leq 0 \\
  -2(x - 1)^2 + 3 & \leq 3 \\
  g(x) & \leq 3
\end{align*}\]
----
\(\{g(x): g(x) \leq 3 \}\)
----
\((-\infty; 3]\)
----
\(y = a(x + p)^2 + q\)
----
\(y = ax^2 + bx +c\)
----
\(y = a(x + p)^2 + q\)
----
\(f(x) = (x-4)^2 - 1\)
----
\[\begin{aligned}
\text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\
\text{Range: } & \left \{ y: y \geq -1, y\in \mathbb{R} \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R} \right \}; \left \{ y: y \geq -1, y\in \mathbb{R} \right \}\)
----
\(g(x) = -(x-5)^2 + 4\)
----
\[\begin{aligned}
\text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\
\text{Range: } & \left \{ y: y \leq 4, y\in \mathbb{R} \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R} \right \}; \left \{ y: y \leq 4, y\in \mathbb{R} \right \}\)
----
\(h(x) = x^2 -6x +9\)
----
\[\begin{aligned}
h(x) &= x^2 -6x +9 \\
    &= (x - 3)^2 \\
\text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\
\text{Range: } & \left \{ y: y \geq 0, y\in \mathbb{R} \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R} \right \}; \left \{ y: y \geq 0, y\in \mathbb{R} \right \}\)
----
\(j(x) = -2(x+1)^2\)
----
\[\begin{aligned}
\text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\
\text{Range: } & \left \{ y: y \leq 0, y\in \mathbb{R} \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R} \right \}; \left \{ y: y \leq 0, y\in \mathbb{R} \right \}\)
----
\(k(x) = -x^2  + 2x - 3\)
----
\[\begin{aligned}
k(x) &= -x^2  + 2x - 3 \\
		 &= -(x - 1)^2 - 2 \\
\text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\
\text{Range: } & \left \{ y: y \leq 2, y\in \mathbb{R} \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R} \right \}; \left \{ y: y \leq 2, y\in \mathbb{R} \right \}\)
----
\(y\)
----
\(y\)
----
\(x\)
----
\(y\)
----
\(x=0\)
----
\(y\)
----
\(g(x) = (x - 1)^2 + 5\)
----
\(x=0\)
----
\[\begin{align*}
	  g(x) &= (x - 1)^2 + 5  \\
	  g(0) &= (0 - 1)^2 + 5 \\ 
	  &= 6
	  \end{align*}\]
----
\((0;6)\)
----
\(x\)
----
\(x\)
----
\(y\)
----
\(x\)
----
\(y=0\)
----
\(x\)
----
\(g(x) = (x - 1)^2 + 5\)
----
\(y=0\)
----
\[\begin{align*}
	  g(x) &= (x - 1)^2 + 5  \\
	  0 &= (x - 1)^2 + 5  \\
	  -5  &= (x - 1)^2
	  \end{align*}\]
----
\(g(x)\)
----
\(x\)
----
\(x\)
----
\(x\)
----
\(y\)
----
\(f(x) = (x+4)^2 - 1\)
----
\[\begin{aligned}
\text{For } x=0 \quad y &= (0+4)^2 - 1 \\
&=  16 - 1 \\
\therefore & (0;15) \\
\text{For } y=0 \quad 0 &= (x+4)^2 - 1 \\
&= x^2 + 8x + 16 - 1 \\
&= x^2 + 8x + 15 \\
&= (x + 5)(x + 3) \\
x=-5 &\text{ or } x=-3 \\
\therefore  (-5;0) &\text{ and } (-3;0) 
\end{aligned}\]
----
\((0;15)\)
----
\((-5;0); (-3;0)\)
----
\[\begin{aligned}
\text{For } x=0 \quad y &=16 \\
\therefore & (0;16) \\
\text{For } y=0 \quad 0 &= 16 - 8x + x^2 \\
&= x^2 - 8x + 16  \\
&= (x -4)^2 \\
\therefore x &=4 \\
\therefore   & (4;0) 
\end{aligned}\]
----
\((0;16)\)
----
\((4;0)\)
----
\(h(x) = -x^2 +4x-3\)
----
\[\begin{aligned}
\text{For } x=0 \quad y &=-3 \\
\therefore & (0;-3) \\
\text{For } y=0 \quad 0 &= -x^2 +4x-3 \\
&= - (x^2 -4x+3) \\
&= -(x - 3)(x - 1) \\
x=3 &\text{ or } x=1 \\
\therefore  (1;0) &\text{ and } (3;0) 
\end{aligned}\]
----
\((0;-3)\)
----
\((1;0); (3;0)\)
----
\(j(x) = 4(x-3)^2 -1\)
----
\[\begin{aligned}
\text{For } x=0 \quad y &= 4(0-3)^2 -1 \\
&= 36 - 1 \\
\therefore & (0;35) \\
\text{For } y=0 \quad 0 &= 4(x-3)^2 -1 \\
&= 4(x^2 - 6x + 9) -1 \\
&= 4x^2 -24x + 36 - 1 \\
&= 4x^2 -36x + 35 \\
&= (2x + 5)(2x + 7) \\
x=-\frac{5}{2} &\text{ or } x=-\frac{7}{2}  \\
\therefore  (-\frac{5}{2};0) &\text{ and } (-\frac{7}{2};0) 
\end{aligned}\]
----
\((0;35)\)
----
\((-\frac{7}{2};0); (-\frac{5}{2};0)\)
----
\(k(x) = 4(x-3)^2 +1\)
----
\[\begin{aligned}
\text{For } x=0 \quad y &= 4(0-3)^2 +1 \\
&= 36 +1 \\
\therefore & (0;37) \\
\text{For } y=0 \quad 0 &= 4(x-3)^2 +1 \\
&= 4(x^2 - 6x + 9) +1 \\
&= 4x^2 -24x + 36 + 1 \\
&= 4x^2 -36x + 37 \\
\therefore x & = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
& = \frac{24 \pm \sqrt{(-24)^2 - 4(4)(37)}}{2(4)} \\
& = \frac{576 \pm \sqrt{576 - 592}}{8} \\
& = \frac{576 \pm \sqrt{-16}}{8} \\
\therefore &\text{ no real solution }
\end{aligned}\]
----
\((0;37)\)
----
\(x\)
----
\(l(x) = 2x^2 - 3x -4\)
----
\((0;-4)\)
----
\(f(x) = a(x+p)^2 + q\)
----
\(a > 0\)
----
\(f(x)\)
----
\([q;\infty)\)
----
\(f(x)\)
----
\(q\)
----
\(f(x) = q\)
----
\(a(x+p)^2 = 0\)
----
\(x = -p\)
----
\((-p;q)\)
----
\(a < 0\)
----
\(f(x)\)
----
\((-\infty;q]\)
----
\(f(x)\)
----
\(q\)
----
\(f(x) = q\)
----
\(a(x+p)^2 = 0\)
----
\(x = -p\)
----
\((-p;q)\)
----
\(f(x) = a(x+p)^2 + q\)
----
\((-p;q)\)
----
\[y = a(x - p)^2 +q\]
----
\(p\)
----
\(p>0\)
----
\(p\)
----
\(p<0\)
----
\(p\)
----
\((p;q)\)
----
\(x = p\)
----
\(g(x) = 3x^2 - 6x - 1\)
----
\(y = a(x+p)^2 + q\)
----
\[\begin{align*}
		  g(x )&= 3x^2 - 6x - 1 \\
		  &= 3(x^2 - 2x) - 1   \\
		  &= 3 \left( (x-1)^2 - 1 \right) -1  \\
		  &= 3(x-1)^2 - 3 -1  \\
		  &= 3(x-1)^2 - 4
		  \end{align*}\]
----
\((-p;q)\)
----
\(g(x) = 3(x-1)^2 - 4\)
----
\(g(x)\)
----
\((1;-4)\)
----
\(x\)
----
\(h(x) = ax^2 + bx + c\)
----
\(x = -\frac{b}{2a}\)
----
\(k(x) = 2 - 10x + 5x^2\)
----
\(y = a(x+p)^2 + q\)
----
\(p = \frac{b}{2a}\)
----
\[\begin{align*}
		    h(x)&= ax^2 + bx + c \\
		    &= a \left( x^2 + \frac{b}{a}x +  \frac{c}{a} \right)
		    \end{align*}\]
----
\(x\)
----
\[\begin{align*}
		    h(x) &= a \left( x^2 + \frac{b}{a}x + \left( \frac{b}{2a} \right)^2 - \left( \frac{b}{2a} \right)^2 +  \frac{c}{a} \right)   \\
		    &= a \left(  \left(x + \frac{b}{2a} \right)^2 - \frac{b^2}{4a^2}  +  \frac{c}{a} \right)   \\
		    &= a \left(  \left(x + \frac{b}{2a} \right)^2 - \frac{b^2 -4ac}{4a^2} \right)   \\
		    &= a  \left(x + \frac{b}{2a} \right)^2 - \frac{b^2 -4ac}{4a}
		    \end{align*}\]
----
\(x = -p = - \frac{b}{2a}\)
----
\(y = q = - \frac{b^2 -4ac}{4a}\)
----
\(k(x)\)
----
\(y = ax^2 + bx + c\)
----
\[k(x) = 5x^2 -10x + 2\]
----
\(a = 5\)
----
\(b = -10\)
----
\(c = 2\)
----
\(x = - \frac{b}{2a}\)
----
\[\begin{align*}
		x &= -\left(\frac{-10}{2(5)}\right) \\
		&= 1
		\end{align*}\]
----
\(x = 1\)
----
\(y\)
----
\(k(x)\)
----
\((1;-3)\)
----
\(y = x^2 - 6x + 8\)
----
\[\begin{aligned}
y &= x^2 - 6x + 8 \\
 &= (x-3)^{2} - \left( \frac{6}{2} \right)^2 + 8 \\
 &= (x-3)^{2}-1 \\
\therefore \text{turning point }&= (3;-1)
\end{aligned}\]
----
\((3;-1)\)
----
\(y = -x^2 + 4x - 3\)
----
\[\begin{aligned}
y &= -x^2 + 4x - 3 \\
 &= -(x^2 - 4x + 3 \\
 &= - \left( (x-2)^{2} - \left( \frac{4}{2} \right)^2 + 3 \right) \\
 &= -(x-2)^{2}+1 \\
\therefore \text{turning point } &= (2;1)
\end{aligned}\]
----
\((2;1)\)
----
\(y = \frac{1}{2}(x + 2)^2 - 1\)
----
\[\begin{aligned}
y &= \frac{1}{2}(x + 2)^2 - 1 \\
\therefore \text{turning point }&= (-2;-1)
\end{aligned}\]
----
\((-2;-1)\)
----
\(y = 2x^2 + 2x + 1\)
----
\[\begin{aligned}
y &= 2x^2 + 2x + 1 \\
 &= 2 \left( x^2 + x + \frac{1}{2} \right) \\
 &= 2 \left( x+ \frac{1}{2}  \right )^2+\frac{1}{2} \\
\therefore \text{turning point }&= (-\frac{1}{2};\frac{1}{2})
\end{aligned}\]
----
\((-\frac{1}{2};\frac{1}{2})\)
----
\[\begin{aligned}
y &= -3x^2 + 6x + 18 \\
 &= -3(x^2  - 2x - 6) \\
 &= -3 \left((x - 1)^2  - 7 \right) \\
 &= -3(x-1)^2+21 \\
\therefore \text{turning point }&= (1;21)
\end{aligned}\]
----
\((1;21)\)
----
\(y = -2[(x + 1)^2 + 3]\)
----
\[\begin{aligned}
y &= -2[(x + 1)^2 + 3] \\
y &= -2(x + 1)^2 - 6 \\
\therefore \text{turning point }&= (-1;-6)
\end{aligned}\]
----
\((-1;-6)\)
----
\(f(x)=a{\left(x+p\right)}^{2}+q\)
----
\(x=-p\)
----
\((-p;q)\)
----
\(y\)
----
\[\begin{aligned}
y &= 2x^2 - 5x - 18 \\
 &= 2(x^2 - \frac{5}{2}x - 9) \\
 &= 2\left( \left( x - \frac{5}{2} \right)^2 - \frac{25}{16} - 9 \right)  \\
 &= 2\left( \left( x - \frac{5}{2} \right)^2 - \frac{169}{16}\right)  \\
 &= 2\left( x - \frac{5}{4} \right)^2 - \frac{169}{8}  \\
\text{Axis of symmetry: }x & =\frac{5}{4}
\end{aligned}\]
----
\(x =\frac{5}{4}\)
----
\(y = 3(x - 2)^2 + 1\)
----
\[\begin{aligned}
y &= 3(x - 2)^2 + 1 \\
\text{Axis of symmetry: } x & = 2
\end{aligned}\]
----
\(x = 2\)
----
\(y = 4x - x^2\)
----
\[\begin{aligned}
y &= 4x - x^2 \\
 &= -(x^2 - 4x) \\
 &= -(x - 2)^2 + 4  \\
\text{Axis of symmetry: } x & = 2
\end{aligned}\]
----
\(x = 2\)
----
\(y\)
----
\(y = ax^2 + q\)
----
\(f(x)=a{\left(x+p\right)}^{2}+q\)
----
\(f(x)=a{\left(x+p\right)}^{2}+q\)
----
\(a\)
----
\(y\)
----
\(x\)
----
\(y = -\frac{1}{2}(x + 1)^2 - 3\)
----
\(y = a(x + p)^2 + q\)
----
\(a < 0\)
----
\((-p;q)\)
----
\((-1; -3)\)
----
\(x = -p\)
----
\(x = -1\)
----
\(y\)
----
\(y\)
----
\(x = 0\)
----
\[\begin{align*}
		  y  &=  -\frac{1}{2} \left((0) + 1 \right)^2 - 3\\
		  &=  -\frac{1}{2} - 3\\
		  &=  -3\frac{1}{2}
		  \end{align*}\]
----
\((0;-3\frac{1}{2})\)
----
\(x\)
----
\(x\)
----
\(y = 0\)
----
\[\begin{align*}
		  0  &=  -\frac{1}{2} \left(x + 1 \right)^2 - 3\\
		  3 &=  -\frac{1}{2} \left(x + 1 \right)^2  \\
		  -6 &=  \left(x + 1 \right)^2
		  \end{align*}\]
----
\(x\)
----
\(x\)
----
\(\{ x: x \in \mathbb{R} \}\)
----
\(\{ y: y \leq -3, y \in \mathbb{R} \}\)
----
\(y = \frac{1}{2}x^2 - 4x + \frac{7}{2}\)
----
\(y = ax^2 + bx +c\)
----
\(a > 0\)
----
\[a = \frac{1}{2}; \qquad b = -4; \qquad c = \frac{7}{2}\]
----
\(x\)
----
\[\begin{align}
		    x &= -\frac{b}{2a} \\
		    &= -\left(\frac{-4}{2\left( \frac{1}{2} \right)}\right) \\
		    &= 4
		    \end{align}\]
----
\(x = 4\)
----
\(x = 4\)
----
\(y\)
----
\[\begin{align*}
		  y &= \frac{1}{2}x^2 - 4x + \frac{7}{2} \\
		  &= \frac{1}{2}(4)^2 - 4(4) + \frac{7}{2} \\
		  &= 8 -16 +\frac{7}{2} \\
		  &= -4\frac{1}{2}
		  \end{align*}\]
----
\(\left( 4; -4\frac{1}{2} \right)\)
----
\(y\)
----
\(y\)
----
\(x = 0\)
----
\[\begin{align*}
		  y  &=  \frac{1}{2}(0)^2 - 4(0) + \frac{7}{2}\\
		  &=  \frac{7}{2}
		  \end{align*}\]
----
\(\left(0;\frac{7}{2}\right)\)
----
\(x\)
----
\(x\)
----
\(y = 0\)
----
\[\begin{align*}
		    0 &= \frac{1}{2}x^2 - 4x + \frac{7}{2} \\
		    &= x^2 - 8x + 7 \\
		    &= (x - 1)(x - 7)
		    \end{align*}\]
----
\(x = 1\)
----
\(x = 7\)
----
\((1;0)\)
----
\((7;0)\)
----
\(\{ x: x \in \mathbb{R} \}\)
----
\(\{ y: y \geq -4\frac{1}{2}, y \in \mathbb{R} \}\)
----
\(y = 3x^2 + 1\)
----
\(x\)
----
\(x\)
----
\(x + 2\)
----
\(y = 3(x + 2)^2 + 1\)
----
\(x\)
----
\(x - 2\)
----
\(y = 3(x - 2)^2 + 1\)
----
\(m\)
----
\(x\)
----
\((x-m)\)
----
\(m\)
----
\(x\)
----
\((x+m)\)
----
\(n\)
----
\(y\)
----
\((y+n)\)
----
\(n\)
----
\(y\)
----
\((y-n)\)
----
\(y = x^2 - 2x -3\)
----
\(x\)
----
\((x-1)\)
----
\[\begin{align*}
	       y &=  x^2 - 2x -3\\
	       &=  (x-1)^2 - 2(x-1) -3 \\
	       &=  x^2 - 2x + 1 -2x + 2 - 3\\
	       &=  x^2 - 4x
	       \end{align*}\]
----
\(x\)
----
\(x+1\)
----
\(y\)
----
\((y+3)\)
----
\[\begin{align*}
	       y + 3&=  x^2 - 2x -3\\
	       y &=  x^2 - 2x -3 - 3 \\
	       &=  x^2 - 2x -6
	       \end{align*}\]
----
\(y = -x^2 + 4x + 5\)
----
\(\text{Intercepts: } (-1;0),(5;0),(0;5)\)
----
\(\text{Turning point: } (2;9)\)
----
\(\text{Axes of symmetry: } x = 2\)
----
\(\text{Domain: } \{ x: x \in \mathbb{R} \}\)
----
\(\text{Range: } \{ y: y \leq 9, y \in \mathbb{R} \}\)
----
\(y = 2(x + 1)^2\)
----
\(\text{Intercepts: } (-1;0),(0;2)\)
----
\(\text{Turning point: } (-1;0)\)
----
\(\text{Axes of symmetry: } x = -1\)
----
\(\text{Domain: } \{ x: x \in \mathbb{R} \}\)
----
\(\text{Range: } \{ y: y \geq 0, y \in \mathbb{R} \}\)
----
\(y = 3x^2 - 2(x+2)\)
----
\(\text{Domain: } \{ x: x \in \mathbb{R} \}\)
----
\(y = 3(x -2)^2 + 1\)
----
\(\text{Intercepts: } (0;13)\)
----
\(\text{Turning point: } (2;1)\)
----
\(\text{Axes of symmetry: } x = 2\)
----
\(\text{Domain: } \{ x: x \in \mathbb{R} \}\)
----
\(\text{Range: } \{ y: y \geq 1, y \in \mathbb{R} \}\)
----
\(f(x) = -x^2 + 7\)
----
\(g(x) = -(x - 2)^2 + 7\)
----
\(h(x) = (x - 2)^2 - 7\)
----
\(y = ax^2 + bx + c\)
----
\(a > 0\)
----
\(b > 0\)
----
\(c < 0\)
----
\(y = ax^2 + bx + c\)
----
\(a < 0\)
----
\(b = 0\)
----
\(c > 0\)
----
\(y = ax^2 + bx + c\)
----
\(a < 0\)
----
\(b < 0\)
----
\(b^2 - 4ac < 0\)
----
\(y = (x+p)^2 + q\)
----
\(p < 0\)
----
\(q < 0\)
----
\(x\)
----
\(y = a(x+p)^2 + q\)
----
\(a < 0\)
----
\(p < 0\)
----
\(q > 0\)
----
\(y = a(x+p)^2 + q\)
----
\(a > 0\)
----
\(p = 0\)
----
\(b^2 - 4ac > 0\)
----
\(y = ax^2 + bx + c\)
----
\(y = 2x^2 + 4x + 2\)
----
\[\begin{aligned}
y &=2x^2 + 4x + 2 \\
x &\Rightarrow  x+3 \\
y_{\text{shifted}} &=2(x+3)^2 + 4(x+3) + 2 \\
 &= 2(x^2 + 6x + 9) + 4x+12 + 2 \\
 &= 2x^2 + 12x + 18 + 4x+12 + 2 \\
 &= 2x^2 + 16x + 32 
\end{aligned}\]
----
\(y_{\text{shifted}} = 2x^2 + 16x + 32\)
----
\(y = -(x+1)^2\)
----
\[\begin{aligned}
y &=2x^2 + 4x + 2 \\
y &\Rightarrow  y-1 \\
y_{\text{shifted}} &= -(x+1)^2 + 1 \\
 &= -(x^2 + 2x + 1) + 1 \\
 &= -x^2 - 2x - 1 + 1 \\
 &= -x^2 - 2x
\end{aligned}\]
----
\(y_{\text{shifted}} = -x^2 - 2x\)
----
\(y = 3(x-1)^2 + 2\left(x-\frac{1}{2}\right)\)
----
\[\begin{aligned}
y &= 3(x-1)^2 + 2\left(x-\frac{1}{2}\right) \\
x &\Rightarrow  x-2 \\
y_{\text{shifted}} &= 3(x - 2-1)^2 + 2\left(x - 2 -\frac{1}{2}\right) \\
 &= 3(x - 3)^2 + 2 \left(x - \frac{5}{2}\right) \\
 &= 3(x^2 - 6x + 9) + 2x - 5 \\
 &= 3x^2 - 18x + 27 + 2x - 5 \\
 &= 3x^2 - 16x + 22
\end{aligned}\]
----
\(y_{\text{shifted}} = 3x^2 - 16x + 22\)
----
\(y = a(x - x_1)(x - x_2)\)
----
\(x\)
----
\((-1;0)\)
----
\((4;0)\)
----
\(\begin{aligned}
		      y &= a(x -x_1)(x - x_2) \\
		      &=  a(x + 1)(x - 4) \\
		      &=  ax^2 - 3ax - 4a
		      \end{aligned}\)
----
\(y\)
----
\((0;2)\)
----
\(\begin{aligned}
		      -4a  &=  2 \\
		      a &=  -\frac{1}{2}
		      \end{aligned}\)
----
\(\begin{aligned}
		      y &=  ax^2 - 3ax - 4a \\
		      &=  -\frac{1}{2}x^2 - 3\left(-\frac{1}{2}\right)x - 4\left(-\frac{1}{2}\right) \\
		      &=  -\frac{1}{2}x^2 + \frac{3}{2}x + 2
		      \end{aligned}\)
----
\(x\)
----
\(y = a(x -x_1)(x - x_2)\)
----
\(x\)
----
\((1;0)\)
----
\((5;0)\)
----
\(\begin{aligned}
		      y  &=  a(x -x_1)(x - x_2) \\
		      &=  a(x - 1)(x - 5) \\
		      &=  ax^2 - 6ax + 5a
		      \end{aligned}\)
----
\(\begin{aligned}
		      y &=  ax^2 - 6ax + 5a \\
		      &=  x^2 - 6x + 5
		      \end{aligned}\)
----
\(y = a(x + p)^2 + q\)
----
\((-3;1)\)
----
\(\begin{aligned}
		      y  &=   a(x + p)^2 + q \\
		      &=   a(x + 3)^2 + 1 \\
		      &=   ax^2 + 6ax + 9a + 1
		      \end{aligned}\)
----
\((1;5)\)
----
\(\begin{aligned}
		      5  &=  a(1)^2 + 6a(1) + 9a + 1 \\
		      4  &=  16a \\
		      \frac{1}{4} &=  a
		      \end{aligned}\)
----
\(y = \frac{1}{4}(x + 3)^2 + 1\)
----
\(y = a(x + p)^2 + q\)
----
\[\begin{aligned}
y &= a(x + p)^2 + q \\
\text{Subst. } & (-1;6) \\
 y &=a(x+1)^2+6 \\
  &=ax^2+2ax+a+6 \\
y-\text{int: } &= (0;3) \\
 3 &=0 +0 +a+6 \\
\therefore 3 &= a + 6 \\
a &= -3 \\
y=-3(x+1)^2+6 &\text{ or } y =-3x^2-6x+3
\end{aligned}\]
----
\(y=-3(x+1)^2+6 \text{ or } y =-3x^2-6x+3\)
----
\[\begin{aligned}
y &= a(x + p)^2 + q \\
y&=a(x-0)(x-5) \\
&=ax^2-5ax \\
\text{Subst. } & (-1;3) \\
 y &=ax^2-5ax \\
 3 &=a(-1)^2-5a(-1) \\
3 &= a+5a \\
3 &= 6a \\
\therefore a &= \frac{1}{2} \\
\therefore y&=\frac{1}{2}x^2-\frac{5}{2}x
\end{aligned}\]
----
\(y=\frac{1}{2}x^2-\frac{5}{2}x\)
----
\[\begin{aligned}
y &= a(x + p)^2 + q \\
\text{Subst. } &= (-2;0) \\
y &= a(x + 2)^2  \\
  &=ax^2+4ax+4a \\
\text{Subst. } & (1;6) \\
 6 &= a +4a +4a \\
 6 &=9a \\
\therefore a &=\frac{2}{3} \\
\therefore y &= \frac{2}{3}(x+2)^2
\end{aligned}\]
----
\(y= \frac{2}{3}(x+2)^2\)
----
\[\begin{aligned}
y &= ax^2+bx+c \\
y &=ax^2+bx+4 \\
\text{Subst. }(1;6): \qquad 6&=a+b+4  \ldots (1) \\
\text{Eqn. }(1) \times 5: \qquad 30 &=5a+5b+20  \ldots (3)  \\   
\text{Subst. }(1;6): \qquad -6&=25a+5b+4  \ldots (2) \\
(2) - (3) \quad -36&=20a-16     \\
20a&=-20 \\
a &= -1 \\
b&=3 \\
c&= 4 \\
\therefore y&=-x^2+3x+4
\end{aligned}\]
----
\(y=-x^2+3x+4\)
----
\(AC\)
----
\[\begin{align*}
	  \text{Gradient }&= \frac{y_A - y_C}{x_A - x_C}  \\
	  &= \frac{7-(-1)}{-3-(-1)} \\
	  &= \frac{8}{-2} \\
	  &= -4
	  \end{align*}\]
----
\(A\)
----
\(C\)
----
\(y=-2{x}^{2}-5\)
----
\(B\)
----
\(\left(0;-5\right)\)
----
\(A\)
----
\(y\)
----
\(A\)
----
\(x\)
----
\(A\)
----
\(B\)
----
\(x_A\)
----
\(y_A\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(C\)
----
\(A\)
----
\(C\)
----
\(A\)
----
\(C\)
----
\(P\left(a;g(a)\right)\)
----
\(Q\left(a+h;g(a+h)\right)\)
----
\(g(x)={x}^{2}\)
----
\(P\left(2;g(2)\right)\)
----
\(Q\left(5;g(5)\right)\)
----
\(Q\)
----
\(P\)
----
\(x\)
----
\[\begin{align*}
		    x_1 &= a \\
		    x_2 &= a + h 
		  \end{align*}\]
----
\(y\)
----
\(g(x) = x^2\)
----
\[y_1 = g(a) = a^2\]
----
\[\begin{align*}
		    y_2 &= g(a + h) \\
		    &= (a + h)^2 \\
		    &= a^2 + 2ah + h^2
		  \end{align*}\]
----
\[\begin{align}
  \frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}
  &= \frac{\left({a}^{2}+2ah+{h}^{2}\right)-\left({a}^{2}\right)}{\left(a+h\right)-\left(a\right)} \\
  &= \frac{{a}^{2}+2ah+{h}^{2}-{a}^{2}}{a+h-a} \\
  &= \frac{2ah+{h}^{2}}{h} \\
  &= \frac{h(2a+h)}{h} \\
  &= 2a+h
\end{align}\]
----
\(P\left(a;g(a)\right)\)
----
\(Q\left(a+h;g(a+h)\right)\)
----
\(g(x)={x}^{2}\)
----
\(2a+h\)
----
\(P\left(2;g(2)\right)\)
----
\(Q\left(5;g(5)\right)\)
----
\(x\)
----
\(P\)
----
\(a\)
----
\(x\)
----
\(Q\)
----
\(a+h\)
----
\(a=2\)
----
\(a+h=5\)
----
\(h=3\)
----
\(2a+h=2\left(2\right)+\left(3\right)=7\)
----
\(Q\)
----
\(P\)
----
\(Q\)
----
\(P\)
----
\(h\)
----
\(Q\)
----
\(P\)
----
\(h=0\)
----
\(2a\)
----
\(f(x)\)
----
\(P\)
----
\(Q\)
----
\(P\left(a;f(a)\right)\)
----
\(Q\left(a+h;f(a+h)\right)\)
----
\(P\)
----
\(Q\)
----
\[\begin{align}
  \text{Average gradient }&= \frac{{y}_{Q}-{y}_{P}}{{x}_{Q}-{x}_{P}} \\
  &= \frac{f(a+h)-f(a)}{\left(a+h\right)-\left(a\right)} \\
  &= \frac{f(a+h)-f(a)}{h}
\end{align}\]
----
\(f(x) = -2x^2\)
----
\(A\)
----
\(x = 1\)
----
\(B\)
----
\(x = 3\)
----
\(A\)
----
\(a < 0\)
----
\(x = 0\)
----
\(y = 0\)
----
\(A\)
----
\(B\)
----
\[\begin{align*}
		  \text{Average gradient} &= \frac{f(3) - f(1)}{3 - 1} \\
		  &= \frac{-2(3)^2 - (-2(1)^2)}{2} \\
		  &= \frac{-18 + 2}{2} \\
		  &= \frac{-16}{2} \\
		  &= -8
		  \end{align*}\]
----
\(f(x)\)
----
\[\begin{align*}
		    \text{Average gradient} &= \frac{f(a + h) - f(a)}{(a + h) - a} \\
		    &= \frac{-2(a + h)^2 - (-2a^2)}{h} \\
		    &= \frac{-2a^2 - 4ah - 2h^2 + 2a^2}{h} \\
		    &= \frac{-4ah - 2h^2}{h} \\
		    &= \frac{h(-4a - 2h)}{h} \\
		    &= -4a - 2h
		    \end{align*}\]
----
\(A\)
----
\(h = 0\)
----
\(a = 1\)
----
\[\begin{align*}
		    \text{Average gradient} &= -4a - 2h \\
		    &= -4(1) - 2(0) \\
		    &= -4
		    \end{align*}\]
----
\(f(x)=x\left(x+3\right)\)
----
\(x=5\)
----
\(x=3\)
----
\(f\)
----
\(x=5\)
----
\(x=3\)
----
\(A\left(1;3\right)\)
----
\(f(x)=3{x}^{2}\)
----
\(f(x)\)
----
\(A\)
----
\(A\)
----
\(A\)
----
\(y = 6x - 3\)
----
\(g(x) = -x^2 + 1\)
----
\(g(x)\)
----
\(x=-2\)
----
\(x=1\)
----
\(g\)
----
\(x=2\)
----
\(g\)
----
\(x=0\)
----
\(y=\frac{a}{x}+q\)
----
\(y = \frac{a}{x} + q\)
----
\(a\)
----
\(q\)
----
\(a\)
----
\(q\)
----
\(f(x) = \frac{a}{x} + q\)
----
\(q\)
----
\(q>0\)
----
\(f(x)\)
----
\(q\)
----
\(q<0\)
----
\(f(x)\)
----
\(q\)
----
\(y = q\)
----
\(y\)
----
\(x = 0\)
----
\(a\)
----
\(a>0\)
----
\(f(x)\)
----
\(a > 1\)
----
\(f(x)\)
----
\(y = \frac{1}{x}\)
----
\(0<a<1\)
----
\(a\)
----
\(f(x)\)
----
\(y = \frac{1}{x}\)
----
\(a<0\)
----
\(f(x)\)
----
\(a < -1\)
----
\(f(x)\)
----
\(y = - \frac{1}{x}\)
----
\(-1<a<0\)
----
\(a\)
----
\(f(x)\)
----
\(y = -\frac{1}{x}\)
----
\(a<0\)
----
\(a>0\)
----
\(q>0\)
----
\(q=0\)
----
\(q<0\)
----
\(y_1 = \frac{1}{x}\)
----
\(y_2 = -\frac{4}{x}\)
----
\(y_3 = \frac{4}{x} - 2\)
----
\(y_4 = -\frac{4}{x} + 1\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(q\)
----
\(q = 0\)
----
\(q\)
----
\(a\)
----
\(a = 1\)
----
\(a\)
----
\(y\)
----
\(x = 0\)
----
\(x\)
----
\(y = 0\)
----
\(y = x\)
----
\(y = -x\)
----
\(\{x: x \in \mathbb{R}, x \ne 0 \}\)
----
\(\{y: y \in \mathbb{R}, y \ne 0 \}\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(q\)
----
\(q = 0\)
----
\(q = 0\)
----
\(q = -2\)
----
\(q = 1\)
----
\(q\)
----
\(a\)
----
\(a = 1\)
----
\(a = -1\)
----
\(a = 4\)
----
\(a = -4\)
----
\(a\)
----
\(y\)
----
\(x = 0\)
----
\(x\)
----
\(y = 0\)
----
\(y\)
----
\(x = 0\)
----
\(x\)
----
\(y = 0\)
----
\(y\)
----
\(x = 0\)
----
\(y = -2\)
----
\(y\)
----
\(x = 0\)
----
\(y = 1\)
----
\(y = x\)
----
\(y = -x\)
----
\(y = x\)
----
\(y = -x\)
----
\(y = x - 2\)
----
\(y = -x - 2\)
----
\(y = x + 1\)
----
\(y = -x + 1\)
----
\(\{x: x \in \mathbb{R}, x \ne 0 \}\)
----
\(\{x: x \in \mathbb{R}, x \ne 0 \}\)
----
\(\{x: x \in \mathbb{R}, x \ne 0 \}\)
----
\(\{x: x \in \mathbb{R}, x \ne 0 \}\)
----
\(\{y: y \in \mathbb{R}, y \ne 0 \}\)
----
\(\{y: y \in \mathbb{R}, y \ne 0 \}\)
----
\(\{y: y \in \mathbb{R}, y \ne -2 \}\)
----
\(\{y: y \in \mathbb{R}, y \ne 1 \}\)
----
\(y=\frac{a}{x+p}+q\)
----
\(y=\frac{a}{x+p}+q\)
----
\(p\)
----
\(a\)
----
\(p\)
----
\(q\)
----
\(y_1 = \frac{1}{x}\)
----
\(y_2 = \frac{1}{x-2}\)
----
\(y_3 = \frac{1}{x-1}\)
----
\(y_4 = \frac{1}{x+1}\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(p\)
----
\(y = \frac{a}{x + p} + q\)
----
\(p\)
----
\(\ldots \ldots\)
----
\(p\)
----
\(\ldots \ldots\)
----
\(q\)
----
\(\ldots \ldots\)
----
\(p\)
----
\(\ldots \ldots\)
----
\(y = \frac{a}{x+p} + q\)
----
\(p\)
----
\(p>0\)
----
\(p\)
----
\(p<0\)
----
\(p\)
----
\(p\)
----
\(x = -p\)
----
\(q\)
----
\(q\)
----
\(y = q\)
----
\(a\)
----
\(p>0\)
----
\(p<0\)
----
\(a<0\)
----
\(a>0\)
----
\(a<0\)
----
\(a>0\)
----
\(q>0\)
----
\(q<0\)
----
\(f(x) = y = \frac{a}{x+p} + q\)
----
\(\{ x: x \in \mathbb{R}, x \ne -p \}\)
----
\(x = -p\)
----
\[y = \frac{a}{x+p} + q\]
----
\[y-q = \frac{a}{x+p}\]
----
\(x \ne -p\)
----
\[\begin{align}
	        \left(y-q\right)\left(x+p\right) &= a \\
	        x + p &= \frac{a}{y-q}
	      \end{align}\]
----
\(\{ y: y \in \mathbb{R}, y \ne q \}\)
----
\(x=-p\)
----
\(y=q\)
----
\(g(x) = \frac{2}{x+1} + 2\)
----
\(\{x: x \in \mathbb{R}, x \ne -1 \}\)
----
\(g(x)\)
----
\(x = -1\)
----
\(g(x) = y\)
----
\[\begin{align*}
		    y &= \frac{2}{x+1} + 2 \\
		    y - 2 &= \frac{2}{x+1} \\
		    (y-2)(x+1) &= 2 \\
		    x + 1 &= \frac{2}{y-2}
		    \end{align*}\]
----
\(\{g(x): g(x) \in \mathbb{R}, g(x) \ne 2 \}\)
----
\(y = \frac{1}{x} + 1\)
----
\[\begin{aligned}
\text{Domain: } & \left \{ x: x \in \mathbb{R},x \neq 0  \right \} \\
\text{Range: } & \left \{ y: y \in \mathbb{R},y \neq 1 \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R},x \neq 0  \right \}; \left \{ y: y \in \mathbb{R},y \neq 1 \right \}\)
----
\(g(x) = \frac{8}{x - 8} +4\)
----
\[\begin{aligned}
\text{Domain: } & \left \{ x: x \in \mathbb{R},x \neq 8  \right \} \\
\text{Range: } & \left \{ y: y \in \mathbb{R},y \neq 4 \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R},x \neq 8  \right \}; \left \{ y: y \in \mathbb{R},y \neq 4 \right \}\)
----
\(y = -\frac{4}{x + 1} -3\)
----
\[\begin{aligned}
\text{Domain: } & \left \{ x: x \in \mathbb{R},x \neq -1  \right \} \\
\text{Range: } & \left \{ y: y \in \mathbb{R},y \neq -3 \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R},x \neq -1  \right \}; \left \{ y: y \in \mathbb{R},y \neq -3 \right \}\)
----
\(x = \frac{2}{3 - y} + 5\)
----
\[\begin{aligned}
x &= \frac{2}{3 - y} + 5 \\
x -5 &= \frac{2}{3 - y} \\
(x -5)(3 - y) &= 2\\
3 - y &= \frac{2}{3x - 5} \\
 - y &= \frac{2}{x - 5} - 3 \\
 \therefore y &= -\frac{2}{x - 5} + 3 \\
\text{Domain: } & \left \{ x: x \in \mathbb{R},x \neq 5  \right \} \\
\text{Range: } & \left \{ y: y \in \mathbb{R},y \neq 3 \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R},x \neq 5 \right \}; \left \{ y: y \in \mathbb{R},y \neq 3 \right \}\)
----
\((y - 2)(x + 2) = 3\)
----
\[\begin{aligned}
(y - 2)(x + 2) &= 3\\
  y - 2 &= \frac{3}{x + 2} \\
 \therefore y   &= \frac{3}{x + 2} + 2 \\
\text{Domain: } & \left \{ x: x \in \mathbb{R},x \neq -2  \right \} \\
\text{Range: } & \left \{ y: y \in \mathbb{R},y \neq 2 \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R},x \neq -2 \right \}; \left \{ y: y \in \mathbb{R},y \neq 2 \right \}\)
----
\(y\)
----
\(y\)
----
\(x=0\)
----
\(y\)
----
\(g(x) = \frac{2}{x + 1} + 2\)
----
\(x=0\)
----
\[\begin{align*}
	     g(x) &= \frac{2}{x + 1} + 2 \\
	     g(0) &= \frac{2}{0 + 1} + 2\\ 
	     &= 2 + 2 \\ 
	     &= 4
	     \end{align*}\]
----
\((0;4)\)
----
\(x\)
----
\(x\)
----
\(y=0\)
----
\(x\)
----
\(g(x) = \frac{2}{x + 1} + 2\)
----
\(y=0\)
----
\[\begin{align*}
	   g(x) &= \frac{2}{x + 1} + 2 \\
	   0 &= \frac{2}{x + 1} + 2 \\
	   -2  &= \frac{2}{x + 1} \\ 
	   -2(x + 1)  &= 2 \\ 
	   -2x - 2  &= 2 \\ 
	   -2x  &= 4 \\ 
	   x  &= -2
	   \end{align*}\]
----
\((-2;0)\)
----
\(x\)
----
\(y\)
----
\(f(x) = \frac{1}{x + 4} - 2\)
----
\[\begin{aligned} 
f(x) &= \frac{1}{x + 4} - 2  \\
\text{Let } x & = 0 \\
f(0) &= \frac{1}{4} - 2 \\
\therefore y_\text{int} &=(0;-1\frac{3}{4}) \\
\text{Let } y & = 0 \\
0 &= \frac{1}{x + 4} - 2  \\
2 &= \frac{1}{x + 4}    \\
2(x + 4) &= 1   \\
2x + 8 &= 1   \\
2x &= -7   \\
\therefore x &= -\frac{7}{2}  \\
\therefore x_\text{int} &= \left( -3\frac{1}{2}; 0 \right) 
\end{aligned}\]
----
\((0;-1\frac{3}{4})\)
----
\(\left( -3\frac{1}{2}; 0 \right)\)
----
\(g(x) = -\frac{5}{x} + 2\)
----
\[\begin{aligned} 
g(x) &= -\frac{5}{x} + 2   \\
\text{Let } x & = 0 \\
\therefore g(x) & \text{is undefined }   \\
\therefore \text{no } & x-\text{intercepts} \\
\text{Let } y & = 0 \\
0 &= -\frac{5}{x} + 2 \\
-2 &= -\frac{5}{x}     \\
2x &= 5  \\
\therefore x &= \frac{5}{2}   \\
\therefore x_\text{int} &= \left( \frac{5}{2} ; 0 \right) 
\end{aligned}\]
----
\(\left( \frac{5}{2} ; 0 \right)\)
----
\(j(x) = \frac{2}{x - 1} + 3\)
----
\[\begin{aligned} 
j(x) &= \frac{2}{x - 1} + 3  \\
\text{Let } x & = 0 \\
j(0) &= \frac{2}{- 1} + 3  \\
&= 1  \\
\therefore y_\text{int} &=(0;1) \\
\text{Let } y & = 0 \\
0 &= \frac{2}{x - 1} + 3 \\
-3 &= \frac{2}{x - 1}     \\
-3(x-1) &= 2  \\
-3x + 3 &= 2   \\
-3x &= -1   \\
\therefore x &= \frac{1}{3}  \\
\therefore x_\text{int} &= \left( \frac{1}{3} ; 0 \right) 
\end{aligned}\]
----
\((0;1)\)
----
\(\left( \frac{1}{3} ; 0 \right)\)
----
\(h(x) = \frac{3}{6 - x} + 1\)
----
\[\begin{aligned} 
h(x) &=  \frac{3}{6 - x} + 1  \\
\text{Let } x & = 0 \\
h(0) &=  \frac{3}{6} + 1  \\
&= \frac{3}{2}   \\
\therefore y_\text{int} &= \left(0;\frac{3}{2} \right) \\
\text{Let } y & = 0 \\
0 &=  \frac{3}{6 - x} + 1 \\
-1 &= \frac{3}{6 - x}     \\
-(6 - x) &= 3  \\
-6 + x &= 3   \\
-3x &= -1   \\
\therefore x &= \frac{1}{3}  \\
\therefore x_\text{int} &= \left( \frac{1}{3} ; 0 \right) 
\end{aligned}\]
----
\(\left(0;\frac{3}{2} \right)\)
----
\(\left( \frac{1}{3} ; 0 \right)\)
----
\(k(x) = \frac{5}{x + 2} - \frac{1}{2}\)
----
\[\begin{aligned} 
k(x) &=  \frac{5}{x + 2} - \frac{1}{2} \\
\text{Let } x & = 0 \\
k(0) &=  \frac{5}{2} - \frac{1}{2} \\
&= 2   \\
\therefore y_\text{int} &= \left(0;2 \right) \\
\text{Let } y & = 0 \\
0 &=  \frac{5}{x + 2} - \frac{1}{2} \\
\frac{1}{2}  &= \frac{5}{x + 2}      \\
x + 2 &= 5(2)  \\
x &= 10 - 2  \\
\therefore x &= 8 \\
\therefore x_\text{int} &= \left( 8 ; 0 \right) 
\end{aligned}\]
----
\(\left(0;2 \right)\)
----
\(\left( 8 ; 0 \right)\)
----
\(y=\frac{a}{x+p}+q\)
----
\(x\)
----
\(y=q\)
----
\(x=-p\)
----
\(y = \frac{1}{x + 4} - 2\)
----
\[\begin{aligned} 
\text{Vertical asymptote: } y &= -2 \\
\text{Horizontal asymptote: } x &= -4 
\end{aligned}\]
----
\(y=-2\)
----
\(x=-4\)
----
\(y = -\frac{5}{x}\)
----
\[\begin{aligned} 
\text{Vertical asymptote: } y &= 0 \\
\text{Horizontal asymptote: } x &= 0 
\end{aligned}\]
----
\(y=0\)
----
\(x=0\)
----
\(y = \frac{3}{2 -x} + 1\)
----
\[\begin{aligned} 
y &= \frac{3}{2 -x} + 1 \\ 
 &= \frac{3}{-(x - 2)} + 1 \\ 
 &= -\frac{3}{x - 2} + 1 \\ 
\text{Vertical asymptote: } y &= 1 \\
\text{Horizontal asymptote: } x &= 2 
\end{aligned}\]
----
\(y=1\)
----
\(x=2\)
----
\(y = \frac{1}{x} -8\)
----
\[\begin{aligned} 
\text{Vertical asymptote: } y &= -8 \\
\text{Horizontal asymptote: } x &= 0 
\end{aligned}\]
----
\(y=-8\)
----
\(x=0\)
----
\(y = -\frac{2}{x - 2}\)
----
\[\begin{aligned} 
\text{Vertical asymptote: } y &= 0 \\
\text{Horizontal asymptote: } x &= 2 
\end{aligned}\]
----
\(y=0\)
----
\(x=2\)
----
\(y =\frac{1}{x}\)
----
\(x \Rightarrow y\)
----
\(y \Rightarrow x\)
----
\(y =\frac{1}{x}\)
----
\(x \Rightarrow -y\)
----
\(y \Rightarrow -x\)
----
\(y = x\)
----
\(y = -x\)
----
\(y =\frac{a}{x + p} + q\)
----
\((-p;q)\)
----
\(y_1 = m_1x + c_1\)
----
\(y_2 = m_2x + c_2\)
----
\(y_1 = x + c_1\)
----
\(y_2 = -x + c_2\)
----
\((-p;q)\)
----
\(c_1\)
----
\(c_2\)
----
\(y = \frac{2}{x + 1} - 2\)
----
\((-p;q)\)
----
\(p = 1\)
----
\(q = -2\)
----
\((-1;-2)\)
----
\[\begin{align*}
		  								y_1 &= x + c_1 \\
		  								y_2 &= -x + c_2
		   								\end{align*}\]
----
\(c_1\)
----
\(c_2\)
----
\((-1;-2)\)
----
\(c_1\)
----
\[\begin{align*}
		  y_1 &= x + c_1 \\
		  -2 &= -1 + c_1 \\
		  -1 &=  c_1
		  \end{align*}\]
----
\((-1;-2)\)
----
\(c_2\)
----
\[\begin{align*}
		  y_2 &= -x + c_2 \\
		  -2 &= -(-1) + c_2 \\
		  -3 &=  c_2
		  \end{align*}\]
----
\(y = \frac{2}{x + 1} - 2\)
----
\[\begin{align*}
		  								y_1 &= x - 1 \\
		  								y_2 &= -x - 3
		   								\end{align*}\]
----
\(f(x)\)
----
\(g(x)\)
----
\((-p;q)\)
----
\(f(x)\)
----
\(g(x)\)
----
\(f(x) = \frac{2}{x}\)
----
\(g(x) = \frac{2}{x} + 1\)
----
\[\begin{aligned} 
(-p;q) &= (0;0) \\
 y_1 &= x  \\
\therefore y_2 &= -x 
\end{aligned}\]
----
\[\begin{aligned} 
(-p;q) &= (0;1) \\
\text{Let } y_1 &= x + c_1 \\
\text{Subst. } (0;1)  \qquad 1 &= 0 + c_1 \\
\therefore c_1 &= 1 \\
\therefore y_1 &= x + 1 \\
\text{Let } y_2 &= -x + c_2 \\
\text{Subst. } (0;1)  \qquad 1 &= 0 + c_2 \\
\therefore c_2 &= 1 \\
\therefore y_2 &= -x + 1 
\end{aligned}\]
----
\(f(x): \quad (0;0); \quad y_1 = x\)
----
\(y_2 = -x\)
----
\(g(x): \quad (0;1); \quad y_1 = x + 1\)
----
\(y_2 = -x + 1\)
----
\(f(x) = -\frac{3}{x}\)
----
\(g(x) = -\frac{3}{x + 1}\)
----
\[\begin{aligned} 
(-p;q) &= (0;0) \\
 y_1 &= x  \\
\therefore y_2 &= -x 
\end{aligned}\]
----
\[\begin{aligned} 
(-p;q) &= (-1;0) \\
\text{Let } y_1 &= x + c_1 \\
\text{Subst. } (-1;0)  \qquad 0 &= -1 + c_1 \\
\therefore c_1 &= 1 \\
\therefore y_1 &= x + 1 \\
\text{Let } y_2 &= -x + c_2 \\
\text{Subst. } (-1;0)  \qquad 0 &= -(-1) + c_2 \\
\therefore c_2 &= -1 \\
\therefore y_2 &= -x - 1 
\end{aligned}\]
----
\(f(x): \quad (0;0); \quad y_1 = x\)
----
\(y_2 = -x\)
----
\(g(x): \quad (-1;0); \quad y_1 = x + 1\)
----
\(y_2 = -x - 1\)
----
\(f(x) = \frac{5}{x}\)
----
\(g(x) = \frac{5}{x - 1} - 1\)
----
\[\begin{aligned} 
(-p;q) &= (0;0) \\
 y_1 &= x  \\
\therefore y_2 &= -x 
\end{aligned}\]
----
\[\begin{aligned} 
(-p;q) &= (1;-1) \\
\text{Let } y_1 &= x + c_1 \\
\text{Subst. } (1;-1)  \qquad -1 &= 1 + c_1 \\
\therefore c_1 &= -2 \\
\therefore y_1 &= x - 2 \\
\text{Let } y_2 &= -x + c_2 \\
\text{Subst. } (1;-1)  \qquad -1 &= -(1) + c_2 \\
\therefore c_2 &= 0 \\
\therefore y_2 &= -x  
\end{aligned}\]
----
\(f(x): \quad (0;0); \quad y_1 = x\)
----
\(y_2 = -x\)
----
\(g(x): \quad (1;-1); \quad y_1 = x -2\)
----
\(y_2 = -x\)
----
\(k(x) = \frac{a}{x +p} + q\)
----
\((4;3)\)
----
\((-1;2)\)
----
\(k(x)\)
----
\[\begin{aligned} 
\text{Given } k(x) &= \frac{a}{x +p} + q \\
\text{and }(-p;q) &= (-1;-2) \\
\therefore &= \frac{a}{x + 1} + 2 \\
\text{Subst. } (4;3)  \qquad 3 &= \frac{a}{4 + 1} + 2 \\
3 - 2 &= \frac{a}{5} \\
\therefore 1(5) &= a \\
\therefore a &= 5 \\
\therefore k(x) &= \frac{5}{x + 1} +  2 
\end{aligned}\]
----
\(k(x) = \frac{5}{x + 1} +  2\)
----
\(f(x)=\frac{a}{x+p}+q\)
----
\(f(x)=\frac{a}{x+p}+q\)
----
\(y\)
----
\(x\)
----
\(y = \frac{2}{x + 1} + 2\)
----
\(y = \frac{a}{x + p} + q\)
----
\(a > 0\)
----
\(p = 1\)
----
\(q = 2\)
----
\(y = 2\)
----
\(x = -1\)
----
\(y\)
----
\(y\)
----
\(x = 0\)
----
\[\begin{align*}
		  y  &=  \frac{2}{0 + 1} + 2 \\
		  &=  4
		  \end{align*}\]
----
\((0;4)\)
----
\(x\)
----
\(x\)
----
\(y = 0\)
----
\[\begin{align*}
		  0  &=  \frac{2}{x + 1} + 2\\
		  -2  &=  \frac{2}{x + 1}  \\
		  -2(x +1) &=  2 \\
		  -2x -2  &=  2 \\
		  -2x &=  4 \\
		  x &= -2
		  \end{align*}\]
----
\((-2;0)\)
----
\((-1;2)\)
----
\(c_1\)
----
\[\begin{align*}
		  y_1 &= x + c_1 \\
		  2 &= -1 + c_1 \\
		  3 &=  c_1
		  \end{align*}\]
----
\[\begin{align*}
		  y_2 &= -x + c_2 \\
		  2 &= -(-1) + c_2 \\
		  1 &=  c_2
		  \end{align*}\]
----
\(y = x + 3\)
----
\(y = -x + 1\)
----
\(\{ x: x \in \mathbb{R}, x \ne -1 \}\)
----
\(\{ y: y \in \mathbb{R}, y \ne 2 \}\)
----
\(f(x) = \frac{1}{x - 2} + 3\)
----
\(y = \frac{a}{x + p} + q\)
----
\(a > 0\)
----
\(y = \frac{1}{x}\)
----
\(g(x) = \frac{1}{x}\)
----
\(x = 0\)
----
\(y = 0\)
----
\(q = 3\)
----
\(g(x)\)
----
\(y = 3\)
----
\(p = -2\)
----
\(g(x)\)
----
\(y\)
----
\(y\)
----
\(x = 0\)
----
\[\begin{align*}
		  y  &=  \frac{1}{0 - 2} + 3 \\
		  &=  2\frac{1}{2}
		  \end{align*}\]
----
\((0;2\frac{1}{2})\)
----
\(x\)
----
\(x\)
----
\(y = 0\)
----
\[\begin{align*}
		  0  &=  \frac{1}{x - 2} + 3\\
		  -3  &=  \frac{1}{x -2}  \\
		  -3(x - 2) &=  1 \\
		  -3x + 6  &=  1 \\
		  -3x &=  -5 \\
		  x &= \frac{5}{3}
		  \end{align*}\]
----
\((\frac{5}{3};0)\)
----
\(\{ x: x \in \mathbb{R}, x \ne 2 \}\)
----
\(\{ y: y \in \mathbb{R}, y \ne 3 \}\)
----
\(a\)
----
\(p\)
----
\(q\)
----
\(y = \frac{a}{x + p} + q\)
----
\(a\)
----
\(a < 0\)
----
\(x = -1\)
----
\(p = 1\)
----
\(y = 3\)
----
\(q = 3\)
----
\[y = \frac{a}{x + 1} + 3\]
----
\(a\)
----
\(a\)
----
\((0;0)\)
----
\[\begin{align*}
		  y  &=  \frac{a}{x + 1} + 3 \\
		  0  &=  \frac{a}{0 + 1} + 3 \\
		  \therefore -3   &=  a
		  \end{align*}\]
----
\[y = -\frac{3}{x + 1} + 3\]
----
\(y = \frac{1}{x} + 2\)
----
\(\text{Asymptotes: } \quad x=0; y=2\)
----
\(\text{Intercepts: } \quad  \left( -\frac{1}{2};0 \right)\)
----
\(\text{Axes of symmetry : } \quad y = x + 2 \text{ and } y = -x + 2\)
----
\(\text{Domain: } \quad  \left \{ x: x \in \mathbb{R},x \neq 0  \right \}\)
----
\(\text{Range: } \quad \left \{ y: y \in \mathbb{R},y \neq 2 \right \}\)
----
\(y = \frac{1}{x + 4} - 2\)
----
\(\text{Asymptotes: } \quad x=-4; y=-2\)
----
\(\text{Intercepts: } \quad  \left( -3\frac{1}{2};0 \right) \text{ and } \left( 0;-1\frac{3}{4} \right)\)
----
\(\text{Axes of symmetry : } \quad y = x + 2 \text{ and } y = -x - 6\)
----
\(\text{Domain: } \quad  \left \{ x: x \in \mathbb{R},x \neq -4  \right \}\)
----
\(\text{Range: } \quad \left \{ y: y \in \mathbb{R},y \neq -2 \right \}\)
----
\(y = -\frac{1}{x + 1} + 3\)
----
\(\text{Asymptotes: } \quad x=-1; y=3\)
----
\(\text{Intercepts: } \quad  \left( -\frac{2}{3};0 \right) \text{ and } \left( 0;2 \right)\)
----
\(\text{Axes of symmetry : } \quad y = x + 4 \text{ and } y = -x +2\)
----
\(\text{Domain: } \quad  \left \{ x: x \in \mathbb{R},x \neq -1  \right \}\)
----
\(\text{Range: } \quad \left \{ y: y \in \mathbb{R},y \neq 3 \right \}\)
----
\(y = -\frac{5}{x - 2\frac{1}{2}} - 2\)
----
\(\text{Asymptotes: } \quad x=-2\frac{1}{2}; y=-2\)
----
\(\text{Intercepts: } \quad  \left( 0;0 \right)\)
----
\(\text{Axes of symmetry : } \quad y = x - 4\frac{1}{2} \text{ and } y = -x + \frac{1}{2}\)
----
\(\text{Domain: } \quad  \left \{ x: x \in \mathbb{R},x \neq -4  \right \}\)
----
\(\text{Range: } \quad \left \{ y: y \in \mathbb{R},y \neq -2 \right \}\)
----
\(y = \frac{8}{x - 8} + 4\)
----
\(\text{Asymptotes: } \quad x=8; y=4\)
----
\(\text{Intercepts: } \quad  \left( 6;0 \right) \text{ and } \left( 0;3 \right)\)
----
\(\text{Axes of symmetry : } \quad y = x - 4 \text{ and } y = -x + 12\)
----
\(\text{Domain: } \quad  \left \{ x: x \in \mathbb{R},x \neq 8  \right \}\)
----
\(\text{Range: } \quad \left \{ y: y \in \mathbb{R},y \neq 4 \right \}\)
----
\(y = \frac{1}{x + p} + q\)
----
\(p\)
----
\(q\)
----
\[\begin{aligned} 
y &= \frac{1}{x + p} + q \\
\text{From graph } \quad p &= 2 \\ 
q &= -1 \\ 
\therefore y &= \frac{1}{x + 2} - 1  
\end{aligned}\]
----
\(y = \frac{1}{x + 2} - 1\)
----
\(y = \frac{a}{x + p} + q\)
----
\(a\)
----
\(p\)
----
\(q\)
----
\[\begin{aligned} 
y &= \frac{a}{x + p} + q \\
\text{From graph } \quad p &= 0 \\ 
q &= 2 \\ 
\therefore y &= \frac{a}{x} + 2 \\
\text{Subst. } (2;0) \quad 0 &= \frac{a}{2} + 2  \\ 
-2 &= \frac{a}{2} \\ 
-2(2) &= a \\ 
\therefore a &= -4 \\
y &= -\frac{4}{x} + 2 
\end{aligned}\]
----
\(y = -\frac{4}{x} + 2\)
----
\(f(x) = -\frac{3}{x}\)
----
\(x > 0\)
----
\(x=1\)
----
\(x=3\)
----
\[\begin{aligned} 
\text{Average gradient } &= \frac{f(3) - f(1)}{3 - 1} \\
&= \dfrac{-\frac{3}{3} - \left( -\frac{3}{1} \right)}{3-1} \\
&= \dfrac{-1 +3}{2} \\
&= \dfrac{2}{2} \\
&= 1
\end{aligned}\]
----
\(x=1\)
----
\(x=3\)
----
\(1\)
----
\(= 1\)
----
\((\frac{1}{2};-6)\)
----
\(x=1\)
----
\(x=3\)
----
\[\begin{aligned} 
\text{Average gradient } &= \frac{f(a+h) - f(a)}{(a+h) - a} \\
&= \dfrac{-\frac{3}{a+h} - \left( -\frac{3}{a} \right)}{(a+h)-a} \\
&= \dfrac{-\frac{3}{a+h}  + \frac{3}{a}}{h} \\
&= \dfrac{\frac{-3(a) + 3(a+h)}{a(a+h)}}{h} \\
&= \dfrac{\frac{-3a + 3a+3h}{a^2+ah}}{h} \\
&= \dfrac{\frac{3h}{a^2+ah}}{h} \\
&= \frac{3h}{a^2+ah} \times \frac{1}{h} \\
&= \frac{3}{a^2+ah} \\
\therefore \text{At } (\frac{1}{2}; -6) \qquad a &= \frac{1}{2} \\
\therefore \text{And  }  \qquad h &= 0 \\
\therefore \text{Average gradient } &= \frac{3}{\left( \frac{1}{2} \right)^2+\frac{1}{2}(0)} \\
&= 3 \times \frac{4}{1} \\
&= 12
\end{aligned}\]
----
\((\frac{1}{2};-6)\)
----
\(x=1\)
----
\(x=3\)
----
\(= 12\)
----
\(y=a{b}^{x}+q\)
----
\(y=a{b}^{x}+q\)
----
\(b>0\)
----
\(a\)
----
\(b\)
----
\(q\)
----
\(a\)
----
\(b\)
----
\(q\)
----
\(f(x) = ab^x + q\)
----
\(q\)
----
\(q>0\)
----
\(f(x)\)
----
\(q\)
----
\(q<0\)
----
\(f(x)\)
----
\(q\)
----
\(y = q\)
----
\(a\)
----
\(a>0\)
----
\(f(x)\)
----
\(a<0\)
----
\(f(x)\)
----
\(b\)
----
\(a > 0\)
----
\(b > 1\)
----
\(f(x)\)
----
\(0 < b < 1\)
----
\(f(x)\)
----
\(b \leq 0\)
----
\(f(x)\)
----
\(b>1\)
----
\(a<0\)
----
\(a>0\)
----
\(q>0\)
----
\(q<0\)
----
\(0<b<1\)
----
\(a<0\)
----
\(a>0\)
----
\(q>0\)
----
\(q<0\)
----
\(y_1 = 3^x\)
----
\(y_2 = -2 \times 3^x\)
----
\(y_3 = 2 \times 3^x + 1\)
----
\(y_4 = 3^x - 2\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(q\)
----
\(q = 0\)
----
\(q\)
----
\(a\)
----
\(a = 1\)
----
\(a\)
----
\(x\)
----
\(y = 0\)
----
\(\{x: x \in \mathbb{R} \}\)
----
\(\{y: y \in \mathbb{R}, y > 0 \}\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(q\)
----
\(q = 0\)
----
\(q = 0\)
----
\(q = 1\)
----
\(q = 2\)
----
\(q\)
----
\(1\)
----
\(2\)
----
\(a\)
----
\(a = 1\)
----
\(a = -2\)
----
\(a = 2\)
----
\(a = 1\)
----
\(a\)
----
\(x\)
----
\(y = 0\)
----
\(x\)
----
\(y = 0\)
----
\(y = 1\)
----
\(y = -2\)
----
\(\{x: x \in \mathbb{R} \}\)
----
\(\{x: x \in \mathbb{R} \}\)
----
\(\{x: x \in \mathbb{R} \}\)
----
\(\{x: x \in \mathbb{R} \}\)
----
\(\{y: y \in \mathbb{R}, y > 0 \}\)
----
\(\{y: y \in \mathbb{R}, y < 0 \}\)
----
\(\{y: y \in \mathbb{R}, y > 1 \}\)
----
\(\{y: y \in \mathbb{R}, y > -2 \}\)
----
\(y=a{b}^{\left(x+p\right)}+q\)
----
\(y=a{b}^{\left(x+p\right)}+q\)
----
\(p\)
----
\(a\)
----
\(p\)
----
\(q\)
----
\(y_1 = 2^x\)
----
\(y_2 = 2^{(x - 2)}\)
----
\(y_3 = 2^{(x - 1)}\)
----
\(y_4 = 2^{(x + 1)}\)
----
\(y_5 = 2^{(x + 2)}\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(y_5\)
----
\(p\)
----
\(y_1 = 2^{(x - 1)} + 2\)
----
\(y_2 = 3 \times 2^{(x - 1)} + 2\)
----
\(y_3 = \frac{1}{2} \times 2^{(x - 1)} + 2\)
----
\(y_4 = 0 \times 2^{(x - 1)} + 2\)
----
\(y_5 = -3 \times 2^{(x - 1)} + 2\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(y_5\)
----
\(a\)
----
\(y = ab^{x + p} + q\)
----
\(p\)
----
\(p>0\)
----
\(p\)
----
\(p<0\)
----
\(p\)
----
\(q\)
----
\(q\)
----
\(y = q\)
----
\(a\)
----
\(a>0\)
----
\(y = q\)
----
\(a<0\)
----
\(y = q\)
----
\(p>0\)
----
\(p<0\)
----
\(a<0\)
----
\(a>0\)
----
\(a<0\)
----
\(a>0\)
----
\(q>0\)
----
\(q<0\)
----
\(f(x) = y = ab^{(x+p)} + q\)
----
\(\left\{x:x\in ℝ\right\}\)
----
\(x\)
----
\(f(x)\)
----
\(f(x)\)
----
\(a\)
----
\(a>0\)
----
\[\begin{align}
  {b}^{\left(x+p\right)} & > 0 \\
  a {b}^{\left(x+p\right)} & > 0 \\
  a {b}^{\left(x+p\right)} + q & > q \\
  f(x) & > q
\end{align}\]
----
\(\{ y: y > q, y \in \mathbb{R} \}\)
----
\(a < 0\)
----
\(\{ y: y < q, y \in \mathbb{R} \}\)
----
\(g(x) = 5 \times 3^{(x+1)} - 1\)
----
\(\{x: x \in \mathbb{R} \}\)
----
\(x\)
----
\(g(x)\)
----
\(g(x)\)
----
\[\begin{align*}
		    3^{(x+1)} & > 0\\
		    5 \times 3^{(x+1)} & > 0\\
		    5 \times 3^{(x+1)} - 1 & > -1\\
		    \therefore g(x) & > -1
		    \end{align*}\]
----
\(\{g(x): g(x) > -1 \}\)
----
\((-1; \infty)\)
----
\(y = \left( \frac{3}{2} \right)^{(x + 3)}\)
----
\[\begin{aligned}
\text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\
\text{Range: } & \left \{ y: y > 0, y\in \mathbb{R} \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R} \right \}; \left \{ y: y > 0, y\in \mathbb{R} \right \}\)
----
\(f(x) =  -5^{(x - 2)} + 1\)
----
\[\begin{aligned}
\text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\
\text{Range: } & \left \{ y: y < 1, y\in \mathbb{R} \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R} \right \}; \left \{ y: y < 1, y\in \mathbb{R} \right \}\)
----
\(y + 3 = 2^{(x + 1)}\)
----
\[\begin{aligned}
y + 3 &= 2^{(x + 1)} \\
y   &= 2^{(x + 1)} - 3 \\
\text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\
\text{Range: } & \left \{ y: y > -3, y\in \mathbb{R} \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R} \right \}; \left \{ y: y > -3, y\in \mathbb{R} \right \}\)
----
\(y = n + 3^{(x - m)}\)
----
\[\begin{aligned}
y &= n + 3^{(x - m)} \\ 
y &= 3^{(x - m)} + n \\ 
\text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\
\text{Range: } & \left \{ y: y > n, y\in \mathbb{R} \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R} \right \}; \left \{ y: y > n, y\in \mathbb{R} \right \}\)
----
\(\frac{y}{2} = 3^{(x - 1)} - 1\)
----
\[\begin{aligned}
\frac{y}{2} &= 3^{(x - 1)} - 1 \\
y &= 2 \times 3^{(x - 1)} - 2 \\
\text{Domain: } & \left \{ x: x \in \mathbb{R} \right \} \\
\text{Range: } & \left \{ y: y > 2, y\in \mathbb{R} \right \}
\end{aligned}\]
----
\(\left \{ x: x \in \mathbb{R} \right \}; \left \{ y: y > 2, y\in \mathbb{R} \right \}\)
----
\(y\)
----
\(y\)
----
\(x=0\)
----
\(y\)
----
\(g(x) = 3 \times 2^{(x + 1)} + 2\)
----
\(x=0\)
----
\[\begin{align*}
	   g(0) &= 3 \times 2^{(0 + 1)} + 2 \\ 
	   &=  3 \times 2 + 2\\ 
	   &=  8
	   \end{align*}\]
----
\((0;8)\)
----
\(x\)
----
\(x\)
----
\(y=0\)
----
\(x\)
----
\(g(x) = 3 \times 2^{(x + 1)} + 2\)
----
\(y=0\)
----
\[\begin{align*}
	   0 &= 3 \times 2^{(x + 1)} + 2  \\
	   -2 &= 3 \times 2^{(x + 1)} \\
	   -\frac{2}{3}  &= 2^{(x + 1)}
	   \end{align*}\]
----
\(g(x)\)
----
\(x\)
----
\(x\)
----
\(x\)
----
\(y\)
----
\(f(x) = 2^{(x + 1)} - 8\)
----
\[\begin{aligned}
\text{For } x=0 \quad y &= 2^{(0 + 1)} - 8 \\
&=  2  - 8 \\
&=  -6 \\
\therefore & (0;-6) \\
\text{For } y=0 \quad 0 &= 2^{(x + 1)} - 8 \\
2^3 &= 2^{(x + 1)} \\
\therefore 3 &= x + 1 \\
\therefore 2 &= x  \\
\therefore  & (2;0) 
\end{aligned}\]
----
\((0;-6)\)
----
\((2;0)\)
----
\[\begin{aligned}
\text{For } x=0 \quad y &= 2 \times 3^{(0 - 1)} - 18 \\
&=  \frac{2}{3}  -18 \\
&=  -17\frac{1}{3} \\
\therefore & (0;-17\frac{1}{3}) \\
\text{For } y=0 \quad 0 &= 2 \times 3^{(x - 1)} - 18 \\
18 &= 2 \times 3^{(x - 1)} \\
9 &=   3^{(x - 1)} \\
3^2 &=   3^{(x - 1)} \\
\therefore 2 &= x - 1 \\
\therefore 3 &= x  \\
\therefore  & (3;0) 
\end{aligned}\]
----
\((0;-17\frac{1}{3})\)
----
\((3;0)\)
----
\(y + 5^{(x + 2)} = 5\)
----
\[\begin{aligned}
y + 5^{(x + 2)} &= 5 \\
y  &= -5^{(x + 2)} + 5 \\
\text{For } x=0 \quad y &= -5^{(0 + 2)} + 5  \\
&=  -25 + 5 \\
&=  -20 \\
\therefore & (0;-20) \\
\text{For } y=0 \quad 0 &= -5^{(x + 2)} + 5  \\
5^{(x + 2)} &= 5\\
\therefore x + 2 &=  1 \\
\therefore x &= -1  \\
\therefore  & (-1;0) 
\end{aligned}\]
----
\((0;-20)\)
----
\((-1;0)\)
----
\((0;\frac{15}{16})\)
----
\((-2;0)\)
----
\(y = ab^{(x+p)} + q\)
----
\(y = q\)
----
\(y = 5 \times 3^{(x+1)} - 1\)
----
\(g(x)\)
----
\[\begin{align*}
		    3^{(x+1)} & \ne 0\\
		    5 \times 3^{(x+1)} & \ne 0\\
		    5 \times 3^{(x+1)} - 1 & \ne -1\\
		    \therefore y & \ne -1
		    \end{align*}\]
----
\(y = -1\)
----
\(y = -5^{(x + 1)}\)
----
\[\begin{aligned} 
y &= -5^{(x + 1)} \\
\text{Horizontal asymptote: } \quad y &= 0 
\end{aligned}\]
----
\(y= 0\)
----
\(y =  3^{(x - 2)} + 1\)
----
\[\begin{aligned} 
y &= 3^{(x - 2)} + 1 \\
\text{Horizontal asymptote: } \quad y &= 1 
\end{aligned}\]
----
\(y= 1\)
----
\(\left( \frac{3y}{2} \right) = 5^{(x + 3)} - 1\)
----
\[\begin{aligned} 
\left( \frac{3y}{2} \right) &= 5^{(x + 3)} - 1 \\
3y &= 2 \times 5^{(x + 3)} - 2 \\
y &= \frac{2}{3} \times 5^{(x + 3)} - \frac{2}{3} \\
\text{Horizontal asymptote: } \quad y &= -\frac{2}{3} 
\end{aligned}\]
----
\(y= -\frac{2}{3}\)
----
\(y = 7^{(x + 1)} - 2\)
----
\[\begin{aligned} 
y &= 7^{(x + 1)} - 2 \\
\text{Horizontal asymptote: } \quad y &= -2 
\end{aligned}\]
----
\(y= -2\)
----
\(\frac{y}{2} + 1 = 3^{(x + 2)}\)
----
\[\begin{aligned} 
\frac{y}{2} + 1 &= 3^{(x + 2)} \\
\frac{y}{2}  &= 3^{(x + 2)} - 1 \\
y  &= 2 \times 3^{(x + 2)} - 2 \\
\text{Horizontal asymptote: }  \quad y &= -2 
\end{aligned}\]
----
\(y= -2\)
----
\(f(x)=a{b}^{\left(x+p\right)}+q\)
----
\(f(x)=a{b}^{\left(x+p\right)}+q\)
----
\(y\)
----
\(x\)
----
\(y = ab^{(x + p)} + q\)
----
\(a > 0\)
----
\(b > 1\)
----
\(y\)
----
\(y\)
----
\(x = 0\)
----
\((0;7\frac{1}{2})\)
----
\(x\)
----
\(x\)
----
\(y = 0\)
----
\((-2;0)\)
----
\(y = -\frac{5}{2}\)
----
\(\{ x: x \in \mathbb{R} \}\)
----
\(\{ y: y > -\frac{5}{2}, y \in \mathbb{R} \}\)
----
\(y = -2 \times 3^{(x + p)} + q\)
----
\(p\)
----
\(q\)
----
\(y = ab^{(x + p)} + q\)
----
\(a = -2\)
----
\(b = 3\)
----
\(p\)
----
\(q\)
----
\(q\)
----
\(y = 6\)
----
\(q = 6\)
----
\[y  = -2 \times 3^{(x + p)} + 6\]
----
\(x\)
----
\(p\)
----
\((2;0)\)
----
\(p\)
----
\[\begin{align*}
		  y  &= -2 \times 3^{(x + p)} + 6 \\
		  0 &= -2 \times 3^{(2 + p)} + 6   \\
		  -6   &= -2 \times 3^{(2 + p)} \\
		  3   &=  3^{(2 + p)}  \\
		  \therefore 1   &= 2 + p \quad \text{(same base)}\\
		  \therefore p   &= -1
		  \end{align*}\]
----
\[y  = -2 \times 3^{(x - 1)} + 6\]
----
\(h(x) = \frac{k}{x}\)
----
\(x < 0\)
----
\(A(-\frac{1}{2}; -6)\)
----
\(k=3\)
----
\[\begin{aligned} 
y &=\frac{k}{x} \\
\text{Subst. } (-\frac{1}{2}; -6) \qquad -6 &=\frac{k}{-\frac{1}{2}} \\
 -6 \times -\frac{1}{2} &= k \\
\therefore k &=3 \\
\therefore h(x) &= \frac{3}{x}
\end{aligned}\]
----
\(h(x)\)
----
\[\begin{aligned} 
h(x) &= \frac{3}{x} \\
\therefore y &\Rightarrow y - 3 \\
y - 3 &=\frac{3}{x}  \\
y  &=\frac{3}{x}+3  
\end{aligned}\]
----
\(y =\frac{3}{x}+3\)
----
\[\begin{aligned} 
h(x) &= \frac{3}{x} \\
\therefore x &\Rightarrow x - 3 \\
y  &=\frac{3}{x -3}  
\end{aligned}\]
----
\(y =\frac{3}{x - 3}\)
----
\(y\)
----
\[\begin{aligned} 
h(x) &= \frac{3}{x} \\
\therefore x &\Rightarrow -x \\
y  &=\frac{3}{x -3}  
\end{aligned}\]
----
\(y =-\frac{3}{x}\)
----
\(x = 0\)
----
\(y = -\frac{1}{4}\)
----
\[\begin{aligned} 
h(x) &= \frac{3}{x} \\
\therefore p=0 &\text{ and } q = -\frac{1}{4} \\
y  &=\frac{3}{x} -\frac{1}{4} 
\end{aligned}\]
----
\(y =\frac{3}{x} -\frac{1}{4}\)
----
\((-1;1)\)
----
\[\begin{aligned} 
h(x) &= \frac{3}{x} \\
\therefore y &\Rightarrow y + m \\
y  &=\frac{3}{x} + m \\
\text{Subst.} (-1;1) \qquad 1 &= \frac{3}{-1} + m \\
1 + 3 &=   + m \\
\therefore m &= 4 \\ 
\therefore y &=\frac{3}{x} + 4 
\end{aligned}\]
----
\(y = \frac{3}{x} + 4\)
----
\(x < 0\)
----
\[\begin{aligned} 
h(x) &= \frac{3}{x} \\
\therefore x &\Rightarrow x + 2 \\
\therefore y &\Rightarrow y + 1 \\
y + 1  &=\frac{3}{x + 2}   \\
\therefore y   &=\frac{3}{x + 2}  - 1 
\end{aligned}\]
----
\(y = \frac{3}{x + 2} - 1\)
----
\(f(x) = a(x+p)^2\)
----
\(g(x) = \frac{a}{x}\)
----
\(f(x)\)
----
\(x = -1\)
----
\(f(x)\)
----
\(g(x)\)
----
\(M\)
----
\(y = 2\)
----
\(M\)
----
\(M\)
----
\(f(x)\)
----
\(x = -1\)
----
\(M(-2;2)\)
----
\(M(-2;2)\)
----
\(g(x)\)
----
\[\begin{aligned} 
 g(x) &=  \frac{a}{x} \\
\text{Subst. } M(-2;2) \qquad 2 &= \frac{a}{-2} \\
\therefore a &= -4 \\
\therefore g(x)&=\frac{-4}{x}
\end{aligned}\]
----
\(g(x) =\frac{-4}{x}\)
----
\(f(x)\)
----
\[\begin{aligned} 
 f(x) &=  a(x + p)^2 + q \\
\text{No vertical shift } \therefore q &= 0 \\
\text{Axis of symmetry } x = -1 \qquad \therefore f(x) &=  a(x + 1)^2  \\
\text{Subst. } M(-2;2) \qquad 2 &=   a(-2 + 1)^2   \\
2 &=   a(-1)^2  \\
\therefore a &= 2 \\
\therefore  f(x) &=  2(x + 1)^2
\end{aligned}\]
----
\(f(x) =  2(x + 1)^2\)
----
\(f(x) < g(x)\)
----
\(- 2 < x < 0\)
----
\(f(x)\)
----
\(\text{Range: }\left \{ y: y\in \mathbb{R}, y \geq  0 \right \}\)
----
\(k(x) = 2(x + \frac{1}{2})^2 - 4\frac{1}{2}\)
----
\(h(x) = 2^{(x + \frac{1}{2})}\)
----
\(\text{For } k(x):\)
----
\(\text{Intercepts: } \quad (-2;0),(1;0) \text{ and } (0;-4)\)
----
\(\text{Turning point: } \quad \left( -\frac{1}{2};-4\frac{1}{2} \right)\)
----
\(\text{Asymptote: } \quad \text{ none }\)
----
\(\text{For } h(x):\)
----
\(\text{Turning point: } \quad \text{ none }\)
----
\(\text{Asymptote: } \quad y=0\)
----
\(h(x)\)
----
\(x\)
----
\(j(x)\)
----
\(y = ax^2 + bx + c\)
----
\(a < 0\)
----
\(b > 0\)
----
\(b^2 < 4ac\)
----
\(a > 0\)
----
\(b > 0\)
----
\(=0\)
----
\(y = \frac{2}{x - 2}\)
----
\(y = \frac{2}{x} - 2\)
----
\(y = -2^{(x - 2)}\)
----
\(f(x) = a(x + p)^2 + q\)
----
\(g(x)=ax^2 + q\)
----
\(h(x) = \frac{a}{x}, x < 0\)
----
\(k(x)=b^x + q\)
----
\[\begin{aligned} 
f(x) &= a(x + p)^2 + q \\
\text{From turning point: } p= -2 &\text{ and } q = 3 \\
\therefore f(x) &= a(x - 2 )^2 + 3 \\
\text{Subst. } (0;0) \qquad 0 &= a(0 - 2)^2 + 3 \\
-3 &= 4a \\
\therefore a &= -\frac{3}{4} \\
f(x) &= -\frac{3}{4}(x - 2)^2 + 3 \\
\text{Axes of symmetry: } x &= 2 \\
\text{Domain: } & \left \{ x:x \in \mathbb{R} \right \} \\
\text{Range: } & \left \{ y:y \in \mathbb{R}, y \leq 3 \right \}
\end{aligned}\]
----
\(f(x)  = -\frac{3}{4}(x - 2)^2 + 3\)
----
\(text{Axes of symmetry: } x  = 2\)
----
\(\text{Domain: }   \{ x:x \in \mathbb{R}  \}\)
----
\(\text{Range: } \left  \{ y:y \in \mathbb{R}, y \leq 3 \right \}\)
----
\[\begin{aligned} 
g(x) &= ax^2 + q \\
\text{From turning point: } p= 0 &\text{ and } q = -2 \\
\therefore g(x) &= a(x)^2 - 2 \\
\text{Subst. } (-2;-1) \qquad -1 &= a(-2 )^2 - 2\\
-1 &= 4a  - 2\\
1 &= 4a  - 2\\
\therefore a &= \frac{1}{4} \\
g(x) &= \frac{1}{4}x^2 -2 \\
\text{Axes of symmetry: } x &= 0 \\
\text{Domain: } & \left \{ x:x \in \mathbb{R} \right \} \\
\text{Range: } & \left \{ y:y \in \mathbb{R}, y \geq -2 \right \} \\
h(x) &= \frac{a}{x + p} + q \\
\text{From graph: } p= 0 &\text{ and } q = 0 \\
h(x) &= \frac{a}{x} \\
\text{Subst. } (-2;-1) \qquad -1 &=\frac{a}{-2} \\
2 &= a \\
\therefore h(x) &= \frac{2}{x} \\
\text{Axes of symmetry: } y &= x \\
\text{Domain: } &\left \{ x:x \in \mathbb{R}, x < 0  \right \} \\
\text{Range: } & \left \{ y:y \in \mathbb{R}, y < 0  \right  \}
\end{aligned}\]
----
\(g(x)=\frac{1}{4}x^2-2\)
----
\(\text{Axes of symmetry: } x = 0\)
----
\(\text{Domain: }   \{ x:x \in \mathbb{R}  \}\)
----
\(\text{Range: }  \{ y:y \in \mathbb{R}, y \geq -2   \}\)
----
\(h(x) = \frac{2}{x}\)
----
\(\text{Axes of symmetry: } y  = x\)
----
\(\text{Domain: }  \left \{ x:x \in \mathbb{R}, x < 0  \right \}\)
----
\(\text{Range: }  \left \{ y:y \in \mathbb{R}, y < 0  \right \}\)
----
\[\begin{aligned} 
y &= 2^{x} + \frac{1}{2} \\
\text{Reflect about } x = 0 \qquad \therefore x &\\Rightarrow -x \\
\therefore k(x) &= 2^{-x} + \frac{1}{2} \\
       &= \left( \frac{1}{2} \right)^{x} + \frac{1}{2} \\
\text{Domain: } & \left \{ x:x \in \mathbb{R} \right \} \\
\text{Range: } & \left \{ y:y \in \mathbb{R}, y > \frac{1}{2} \right \}
\end{aligned}\]
----
\(k(x)  = \left( \frac{1}{2} \right)^{x} + \frac{1}{2}\)
----
\(\text{Domain: } \left  \{ x:x \in \mathbb{R} \right \}\)
----
\(\text{Range: }  \left \{ y:y \in \mathbb{R}, y > \frac{1}{2} \right \}\)
----
\(Q(x) = a^x\)
----
\(a = \frac{1}{3}\)
----
\[\begin{aligned} 
y&=a^x \\
\text{Subst.} \left( 1;\frac{1}{3} \right) \qquad \frac{1}{3} &=a^{1} \\
\therefore a &= \frac{1}{3} 
\end{aligned}\]
----
\(p\)
----
\((-2;p)\)
----
\(Q\)
----
\[\begin{aligned} 
Q(x) &= \left( \frac{1}{3} \right)^x \\
\text{Subst.} \left( -2;p \right) p &= \left( \frac{1}{3} \right)^{-2} \\
 p &= 9 
\end{aligned}\]
----
\(p = 9\)
----
\(x = -2\)
----
\(x = 1\)
----
\[\begin{aligned} 
\text{Average gradient} &= \frac{\left ( \frac{1}{3} \right )^{-2} - \left ( \frac{1}{3} \right )^1}{-2-(1)} \\
&= \frac{9-\frac{1}{3}}{-3} \\
&=\frac{8\frac{2}{3}}{-3} \\
&= \frac{-26}{9} \\
&= -2\frac{8}{9}
\end{aligned}\]
----
\(\text{Average gradient} = -2\frac{8}{9}\)
----
\(Q\)
----
\[\begin{aligned} 
Q(x) &= \left( \frac{1}{3} \right)^x \\
\therefore x & \Rightarrow x + 2 \\
\therefore y & \Rightarrow y + 2 \\
 y + 2 &= \left( \frac{1}{3} \right)^{x + 2} \\
 y   &= \left( \frac{1}{3} \right)^{x + 2} -2 
\end{aligned}\]
----
\(y = \left( \frac{1}{3} \right)^{x + 2} -2\)
----
\(f(x) = 2^x + q\)
----
\(g(x) = mx + c\)
----
\[\begin{aligned} 
f(x) &= 2^x + q \\
\text{Subst. } (0; -\frac{1}{2}) \quad -\frac{1}{2} &= 2^{0} + q  \\
-\frac{1}{2} &= 1 + q \\
\therefore q &= -\frac{3}{2} \\
\therefore f(x) &=2^x-\frac{3}{2} \\
g(x) &= mx +c \\
\text{Subst. } (0; -\frac{1}{2}) \quad -\frac{1}{2} &= m(0) + c \\
\therefore c &= -\frac{1}{2} \\
g(x) &= mx -\frac{1}{2} \\
\text{Subst. } (-2;0) \quad 0 &= m(-2) -\frac{1}{2} \\
 \frac{1}{2} &= -2m  \\
\therefore m &= -\frac{1}{4} \\
\therefore g(x) &= -\frac{1}{4}x-\frac{1}{2}
\end{aligned}\]
----
\(f(x)  = 2^x-\frac{3}{2}\)
----
\(g(x) = -\frac{1}{4}x-\frac{1}{2}\)
----
\(h(x) = \frac{k}{x + p} + q\)
----
\[\begin{aligned} 
h(x) &=  \frac{k}{x + p} + q \\
\text{From graph: } \quad p = -2 \\
h(x) &=  \frac{k}{x + 2} + q \\
\text{Subst. } (0; -\frac{1}{2}) \quad -\frac{1}{2} &= \frac{k}{2} + q  \\
-1 &= k + 2q \ldots (1)\\
\text{Subst. } (1;0) \quad 0 &= \frac{k}{1 + 2} + q  \\
0 &= k + 3q \ldots (2)\\
(2) - (1): \qquad 1 &= 0 + q \\
\therefore q &= 1 \\
\text{and } k &= -3\\
\therefore h(x) &=  -\frac{3}{x + 2} + 1
\end{aligned}\]
----
\(h(x)  =  -\frac{3}{x + 2} + 1\)
----
\(D\)
----
\(h(x) = mx + c\)
----
\(B\)
----
\(C\)
----
\(AO\)
----
\(OB\)
----
\(OC\)
----
\(DE\)
----
\(DE\)
----
\(DE = 12\frac{1}{4}\)
----
\(h(x)\)
----
\[\begin{aligned} 
h(x) & = mx + c \\
h(x) & = mx + 10 \\
\text{Subst. } (5;0) \qquad 0 &= m(5) + 10 \\
-10 &= 5m \\
\therefore m &= -2 \\
\therefore h(x) &=-2x+10
\end{aligned}\]
----
\(h(x) =-2x+10\)
----
\(x\)
----
\(k(x) < 0\)
----
\(\left \{ x: x\in \mathbb{R}, x < -2 \text{ and } x > 5 \right \}\)
----
\(x\)
----
\(k(x) \geq h(x)\)
----
\(\left \{ x: x\in \mathbb{R}, 0 \leq x \leq 5 \right \}\)
----
\(DF\)
----
\(y = \sin \theta\)
----
\(x\)
----
\(y = \sin \theta\)
----
\(\{ \theta: \theta \in \mathbb{R} \}\)
----
\(\left[-1;1\right]\)
----
\(x\)
----
\(y\)
----
\(y = a \sin \theta + q\)
----
\(a\)
----
\(q\)
----
\(f(\theta) = a \sin \theta + q\)
----
\(q\)
----
\(q>0\)
----
\(f(\theta)\)
----
\(q\)
----
\(q<0\)
----
\(f(\theta)\)
----
\(q\)
----
\(a\)
----
\(a>1\)
----
\(f(\theta)\)
----
\(0<a<1\)
----
\(f(\theta)\)
----
\(a<0\)
----
\(x\)
----
\(-1 < a < 0\)
----
\(x\)
----
\(a < -1\)
----
\(x\)
----
\(x\)
----
\(y\)
----
\(y_1 = \sin \theta\)
----
\(y_2 = - 2 \sin \theta\)
----
\(y_3 = \sin \theta + 1\)
----
\(y_4 = \frac{1}{2} \sin \theta - 1\)
----
\(y = \sin k\theta\)
----
\(k\)
----
\(y_1 = \sin \theta\)
----
\(θ\)
----
\(\sin \theta\)
----
\(y_1 = \sin \theta\)
----
\(y_2 = \sin (-\theta)\)
----
\(y_3 = \sin 2\theta\)
----
\(y_4 = \sin \frac{\theta}{2}\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(y\)
----
\(x\)
----
\(k\)
----
\(y_1 = \sin \theta\)
----
\(y_2 = \sin (-\theta)\)
----
\(\sin (-\theta) = -\sin \theta\)
----
\(y = \sin k\theta\)
----
\(y = \sin k\theta\)
----
\(k\)
----
\(k\)
----
\(y\)
----
\(k > 0\)
----
\(k > 1\)
----
\(0 < k < 1\)
----
\(k < 0\)
----
\(-1 < k < 0\)
----
\(y\)
----
\(k < -1\)
----
\(y\)
----
\[\sin (-\theta) = -\sin \theta\]
----
\(y = \sin k\theta\)
----
\(|k|\)
----
\(k\)
----
\(k\)
----
\(0 < k < 1\)
----
\(-1 < k < 0\)
----
\(k > 1\)
----
\(k < -1\)
----
\(y_1 = \sin \theta\)
----
\(y_2 = \sin \frac{3\theta}{2}\)
----
\(x\)
----
\(y\)
----
\(y = \sin k\theta\)
----
\(k > 1\)
----
\(y_2 = \sin \frac{3\theta}{2}\)
----
\(θ\)
----
\(\sin \theta\)
----
\(\sin \frac{3\theta}{2}\)
----
\(y_1 = \sin \theta\)
----
\(y_2 = \sin \frac{3\theta}{2}\)
----
\([-1;1]\)
----
\([-1;1]\)
----
\(y\)
----
\(x\)
----
\(f(\theta) = y =\sin k\theta\)
----
\(\{ \theta: \theta \in \mathbb{R} \}\)
----
\(\theta\)
----
\(f(\theta)\)
----
\(\{ f(\theta): -1 \leq f(\theta) \leq 1, f(\theta) \in \mathbb{R} \}\)
----
\([-1;1]\)
----
\(x\)
----
\(f(\theta) = 0\)
----
\(\theta\)
----
\(y\)
----
\(f(\theta)\)
----
\(y = \sin k\theta\)
----
\(x\)
----
\(y\)
----
\(f(\theta) =\sin 3\theta\)
----
\(f(\theta) =\sin 3\theta\)
----
\(g(\theta) =\sin \frac{\theta}{3}\)
----
\(g(\theta) =\sin \frac{\theta}{3}\)
----
\(h(\theta) = \sin (-2\theta)\)
----
\(h(\theta) =\sin (-2\theta)\)
----
\(k(\theta) =\sin \frac{3\theta}{4}\)
----
\(k(\theta) =\sin \frac{3\theta}{4}\)
----
\(f(\theta) =\sin k\theta\)
----
\(k\)
----
\(y = \sin(\theta + p)\)
----
\(p\)
----
\(y_1 = \sin \theta\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(y_5\)
----
\(y\)
----
\(x\)
----
\(p\)
----
\(y = \sin(\theta + p)\)
----
\(p\)
----
\(p > 0\)
----
\(p\)
----
\(p < 0\)
----
\(p\)
----
\(p>0\)
----
\(p<0\)
----
\(y_1 = \sin \theta\)
----
\(x\)
----
\(y\)
----
\(y = \sin (\theta + p)\)
----
\(y_1 = \sin \theta\)
----
\(p = 0\)
----
\(p < 0\)
----
\(\sin \theta\)
----
\(y_1 = \sin \theta\)
----
\([-1;1]\)
----
\([-1;1]\)
----
\(y\)
----
\(x\)
----
\(f(\theta) = y =\sin (\theta + p)\)
----
\(\{ \theta: \theta \in \mathbb{R} \}\)
----
\(\theta\)
----
\(f(\theta)\)
----
\(\{ f(\theta): -1 \leq f(\theta) \leq 1, f(\theta) \in \mathbb{R} \}\)
----
\(x\)
----
\(f(\theta) = 0\)
----
\(\theta\)
----
\(y\)
----
\(f(\theta)\)
----
\(y = \sin (\theta + p)\)
----
\(x\)
----
\(y\)
----
\(y = \sin (\theta + p)\)
----
\(y = \sin\theta\)
----
\(x\)
----
\(f(\theta)\)
----
\([-1;1]\)
----
\(y\)
----
\(x\)
----
\(y = \sin k(\theta + p)\)
----
\(y = \sin\theta\)
----
\(f(\theta)\)
----
\([-1;1]\)
----
\(y\)
----
\(x\)
----
\(y = 2 \sin \frac{\theta}{2}\)
----
\(y = \sin (-\frac{3\theta}{2})\)
----
\(y = a \sin (\theta + p)\)
----
\(a\)
----
\(p\)
----
\(\cos \theta\)
----
\(a = 2\)
----
\(y = 2 \cos \theta\)
----
\(y = \cos \theta\)
----
\(y = \cos \theta\)
----
\(\{ \theta: \theta \in \mathbb{R} \}\)
----
\(\left[-1;1\right]\)
----
\(x\)
----
\(y\)
----
\(y = a \cos \theta + q\)
----
\(y = a \cos \theta + q\)
----
\(a\)
----
\(q\)
----
\(a\)
----
\(q\)
----
\(f(\theta) = a \cos \theta + q\)
----
\(q\)
----
\(q>0\)
----
\(f(\theta)\)
----
\(q\)
----
\(q<0\)
----
\(f(\theta)\)
----
\(q\)
----
\(a\)
----
\(a>1\)
----
\(f(\theta)\)
----
\(0<a<1\)
----
\(f(\theta)\)
----
\(a<0\)
----
\(x\)
----
\(-1 < a < 0\)
----
\(x\)
----
\(a < -1\)
----
\(x\)
----
\(x\)
----
\(y\)
----
\(y_1 = \cos \theta\)
----
\(y_2 = - 3 \cos \theta\)
----
\(y_3 = \cos \theta + 2\)
----
\(y_4 = \frac{1}{2} \cos \theta - 1\)
----
\(y=\cos (k\theta)\)
----
\(y = \cos k\theta\)
----
\(k\)
----
\(k\)
----
\(y_1 = \cos \theta\)
----
\(\cos \theta\)
----
\(\cos \theta\)
----
\(y_1 = \cos \theta\)
----
\(y_2 = \cos (-\theta)\)
----
\(y_3 = \cos 3\theta\)
----
\(y_4 = \cos \frac{3\theta}{4}\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(y\)
----
\(x\)
----
\(k\)
----
\(y_1 = \cos \theta\)
----
\(y_2 = \cos (-\theta)\)
----
\(\cos (-\theta) = -\cos \theta\)
----
\(y = \cos k\theta\)
----
\(k\)
----
\(y = \cos k\theta\)
----
\(k\)
----
\(k > 0\)
----
\(k > 1\)
----
\(0 < k < 1\)
----
\(k < 0\)
----
\(-1 < k < 0\)
----
\(k < -1\)
----
\[\cos (-\theta) = \cos \theta\]
----
\(\theta\)
----
\(x\)
----
\(y = \cos k\theta\)
----
\(|k|\)
----
\(k\)
----
\(0 < k < 1\)
----
\(-1 < k < 0\)
----
\(k > 1\)
----
\(k < -1\)
----
\(y_1 = \cos \theta\)
----
\(y_2 = \cos \frac{\theta}{2}\)
----
\(x\)
----
\(y\)
----
\(y = \cos k\theta\)
----
\(y_2 = \cos \frac{\theta}{2}\)
----
\(k < 1\)
----
\(\cos \theta\)
----
\(\cos \frac{\theta}{2}\)
----
\(y_1 = \cos \theta\)
----
\(y_2 = \cos \frac{\theta}{2}\)
----
\([-1;1]\)
----
\([0;1]\)
----
\(y\)
----
\(x\)
----
\(f(\theta) = y =\cos k\theta\)
----
\(\{ \theta: \theta \in \mathbb{R} \}\)
----
\(\theta\)
----
\(f(\theta)\)
----
\(\{ f(\theta): -1 \leq f(\theta) \leq 1, f(\theta) \in \mathbb{R} \}\)
----
\([-1;1]\)
----
\(x\)
----
\(f(\theta) = 0\)
----
\(\theta\)
----
\(y\)
----
\(f(\theta)\)
----
\(y = \cos k\theta\)
----
\(x\)
----
\(y\)
----
\(f(\theta) =\cos 2\theta\)
----
\(f(\theta) =\cos 2 \theta\)
----
\(g(\theta) =\cos \frac{\theta}{3}\)
----
\(g(\theta) =\cos \frac{\theta}{3}\)
----
\(h(\theta) =\cos (-2\theta)\)
----
\(h(\theta) =\cos (-2\theta)\)
----
\(k(\theta) =\cos \frac{3\theta}{4}\)
----
\(k(\theta) =\cos \frac{3\theta}{4}\)
----
\(f(\theta) =\cos k\theta\)
----
\(k\)
----
\(k = \frac{3}{2}\)
----
\(k = \frac{2}{3}\)
----
\(y=\cos\left(\theta +p\right)\)
----
\(y = \cos(\theta + p)\)
----
\(p\)
----
\(p\)
----
\(y_1 = \cos \theta\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(y_5\)
----
\(y\)
----
\(x\)
----
\(p\)
----
\(y = \cos(\theta + p)\)
----
\(p\)
----
\(p > 0\)
----
\(p\)
----
\(p < 0\)
----
\(p\)
----
\(p>0\)
----
\(p<0\)
----
\(y_1 = \cos \theta\)
----
\(x\)
----
\(y\)
----
\(y = \cos (\theta + p)\)
----
\(y_1 = \cos \theta\)
----
\(p = 0\)
----
\(p < 0\)
----
\(\cos \theta\)
----
\(y_1\)
----
\(y_2\)
----
\([-1;1]\)
----
\([-1;1]\)
----
\(y\)
----
\(x\)
----
\(f(\theta) = y =\cos (\theta + p)\)
----
\(\{ \theta: \theta \in \mathbb{R} \}\)
----
\(\theta\)
----
\(f(\theta)\)
----
\(\{ f(\theta): -1 \leq f(\theta) \leq 1, f(\theta) \in \mathbb{R} \}\)
----
\(x\)
----
\(f(\theta) = 0\)
----
\(\theta\)
----
\(y\)
----
\(f(\theta)\)
----
\(y = \cos (\theta + p)\)
----
\(x\)
----
\(y\)
----
\(y = \cos k(\theta + p)\)
----
\(y = \cos \theta\)
----
\(f(\theta)\)
----
\([-1;1]\)
----
\(y\)
----
\(x\)
----
\(y = a \cos (k\theta + p)\)
----
\(a\)
----
\(k\)
----
\(p\)
----
\(k\)
----
\(k = 1\)
----
\[y = a \cos ( \theta + p)\]
----
\(a\)
----
\(a = 2\)
----
\[y = 2 \cos ( \theta + p)\]
----
\(p\)
----
\(y = \cos \theta\)
----
\(y = -2\)
----
\(y = -2 \cos \theta\)
----
\(y = a \cos \theta\)
----
\(a\)
----
\(a = -1\)
----
\(y = \cos (\theta + p)\)
----
\(p\)
----
\(y = \tan\theta\)
----
\(\tan\theta\)
----
\(\left\{f(\theta):f(\theta)\in ℝ\right\}\)
----
\(x\)
----
\(y\)
----
\(y = a \tan \theta + q\)
----
\(y = a \tan \theta + q\)
----
\(a\)
----
\(q\)
----
\(a\)
----
\(q\)
----
\(f(\theta) = a \tan \theta + q\)
----
\(q\)
----
\(q>0\)
----
\(f(\theta)\)
----
\(q\)
----
\(q<0\)
----
\(f(\theta)\)
----
\(q\)
----
\(a\)
----
\(a>1\)
----
\(f(\theta)\)
----
\(0<a<1\)
----
\(f(\theta)\)
----
\(a<0\)
----
\(x\)
----
\(-1 < a < 0\)
----
\(x\)
----
\(a < -1\)
----
\(x\)
----
\(a<0\)
----
\(a>0\)
----
\(q>0\)
----
\(q=0\)
----
\(q<0\)
----
\(x\)
----
\(y\)
----
\(y_1 = \tan \theta - \frac{1}{2}\)
----
\(y_2 = - 3 \tan \theta\)
----
\(y_3 = \tan \theta + 2\)
----
\(y_4 = 2 \tan \theta - 1\)
----
\(y=\tan (k\theta)\)
----
\(k\)
----
\(y_1 = \tan \theta\)
----
\(\tan \theta\)
----
\(\tan \theta\)
----
\(y_1 = \tan \theta\)
----
\(y_2 = \tan (-\theta)\)
----
\(y_3 = \tan 3\theta\)
----
\(y_4 = \tan \frac{\theta}{2}\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(y\)
----
\(x\)
----
\(k\)
----
\(y_1 = \tan \theta\)
----
\(y_2 = \tan (-\theta)\)
----
\(\tan (-\theta) = -\tan \theta\)
----
\(y = \tan k\theta\)
----
\(y = \tan k\theta\)
----
\(k\)
----
\(k\)
----
\(y\)
----
\(k > 0\)
----
\(k > 1\)
----
\(0 < k < 1\)
----
\(k < 0\)
----
\(-1 < k < 0\)
----
\(y\)
----
\(k < -1\)
----
\(y\)
----
\[\tan (-\theta) = -\tan \theta\]
----
\(y = \tan k\theta\)
----
\(|k|\)
----
\(k\)
----
\(k > 0\)
----
\(k < 0\)
----
\(y_1 = \tan \theta\)
----
\(y_2 = \tan \frac{3\theta}{2}\)
----
\(x\)
----
\(y\)
----
\(y = \tan k\theta\)
----
\(k > 1\)
----
\(y_2 = \tan \frac{3\theta}{2}\)
----
\(\tan \theta\)
----
\(\tan \frac{3\theta}{2}\)
----
\(y_1 = \tan \theta\)
----
\(y_2 = \tan \frac{3\theta}{2}\)
----
\(\{f(\theta): f(\theta) \in \mathbb{R}\}\)
----
\(\{f(\theta): f(\theta) \in \mathbb{R}\}\)
----
\(y\)
----
\(x\)
----
\(f(\theta) = y =\tan k\theta\)
----
\(f(\theta)\)
----
\(\{ f(\theta): f(\theta) \in \mathbb{R} \}\)
----
\((-\infty; \infty)\)
----
\(x\)
----
\(f(\theta) = 0\)
----
\(\theta\)
----
\(y\)
----
\(\theta = 0\)
----
\(f(\theta)\)
----
\(k\theta\)
----
\(\tan k\theta\)
----
\(y = \tan k\theta\)
----
\(x\)
----
\(y\)
----
\(f(\theta) =\tan 2\theta\)
----
\(g(\theta) =\tan \frac{3\theta}{4}\)
----
\(h(\theta) =\tan (-2\theta)\)
----
\(k(\theta) =\tan \frac{2\theta}{3}\)
----
\(y=\tan\left(\theta +p\right)\)
----
\(y = \tan(\theta + p)\)
----
\(p\)
----
\(p\)
----
\(y_1 = \tan \theta\)
----
\(y_1\)
----
\(y_2\)
----
\(y_3\)
----
\(y_4\)
----
\(y_5\)
----
\(y\)
----
\(x\)
----
\(p\)
----
\(y = \tan(\theta + p)\)
----
\(p\)
----
\(p > 0\)
----
\(p\)
----
\(p < 0\)
----
\(p\)
----
\(p > 0\)
----
\(p < 0\)
----
\(y_1 = \tan \theta\)
----
\(x\)
----
\(y\)
----
\(y = \tan (\theta + p)\)
----
\(y_1 = \tan \theta\)
----
\(\tan \theta\)
----
\(y_1 = \tan \theta\)
----
\((-\infty;\infty)\)
----
\((-\infty;\infty)\)
----
\(y\)
----
\(x\)
----
\(f(\theta) = y =\tan (\theta + p)\)
----
\(\{ f(\theta): f(\theta) \in \mathbb{R} \}\)
----
\(x\)
----
\(f(\theta) = 0\)
----
\(\theta\)
----
\(y\)
----
\(f(\theta)\)
----
\(y = \tan (\theta + p)\)
----
\(x\)
----
\(y\)
----
\(0 < k < 1\)
----
\(y = \tan \theta\)
----
\(p < 0\)
----
\(x\)
----
\(y = \tan \theta\)
----
\((-\infty;\infty)\)
----
\(y\)
----
\(x\)
----
\(y = \tan \theta - 1\)
----
\(f(\theta) =  -\tan 2\theta\)
----
\(y = a \tan k\theta\)
----
\(a\)
----
\(k\)
----
\(a = -1\)
----
\(k = \frac{1}{2}\)
----
\(f(\theta) = a \sin k\theta\)
----
\(g(\theta) = a \tan \theta\)
----
\(f(\theta) = \frac{3}{2} \sin 2\theta\)
----
\(g(\theta) = -\frac{3}{2} \tan \theta\)
----
\(f(\theta) = a \sin k\theta\)
----
\(g(\theta) = a \cos ( \theta + p)\)
----
\(f(\theta) = -2 \sin \theta\)
----
\(y = a \tan k\theta\)
----
\(y = 3 \tan \frac{\theta}{2}\)
----
\(y = a \cos \theta + q\)
----
\(y = y = 2 \cos \theta + 2\)
----
\(f(\theta) = 2 \sin \theta\)
----
\(g(\theta) = \cos \theta + 1\)
----
\(f\)
----
\(g\)
----
\(2 \sin \theta - \cos \theta = 1\)
----
\(2 \sin \theta = -1\)
----
\(f(\theta) = a \cos \theta\)
----
\(g(\theta) = \tan \theta\)
----
\(Q(c; -1)\)
----
\(g(\theta)\)
----
\(f(\theta)\)
----
\(a\)
----
\(b\)
----
\(c\)
----
\(a = 2\)
----
\(b = -1\)
----
\(g\)
----
\(\cos \theta = \frac{1}{2}\)
----
\(g\)
----
\(x\)
----
\(y = ax^2 + bx + c\)
----
\(y\)
----
\((0;c)\)
----
\(x\)
----
\(x = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\)
----
\(\left(-\frac{b}{2a}; -\frac{b^2}{4a}+c\right)\)
----
\(x = -\frac{b}{2a}\)
----
\(y = a(x+p)^2 + q\)
----
\((-p;q)\)
----
\(p > 0\)
----
\(p < 0\)
----
\(q > 0\)
----
\(q < 0\)
----
\(= \frac{y_2 - y_1}{x_2 - x_1}\)
----
\(y = \frac{k}{x}\)
----
\(k > 0\)
----
\(k < 0\)
----
\(y = \frac{k}{x+p} + q\)
----
\(p > 0\)
----
\(p < 0\)
----
\(q > 0\)
----
\(q < 0\)
----
\(x = -p\)
----
\(y = q\)
----
\(y = ab^x\)
----
\(a > 0\)
----
\(x\)
----
\(a < 0\)
----
\(x\)
----
\(b > 1\)
----
\(a > 0\)
----
\(a < 0\)
----
\(0 < b <  1\)
----
\(a > 0\)
----
\(a < 0\)
----
\(y = ab^{(x +p)} + q\)
----
\(p > 0\)
----
\(p < 0\)
----
\(q > 0\)
----
\(q < 0\)
----
\(y = q\)
----
\(y = a \sin (k \theta + p) + q\)
----
\(k > 1\)
----
\(k < -1\)
----
\(0  <k <1\)
----
\(-1  <k <0\)
----
\(p > 0\)
----
\(p < 0\)
----
\(q > 0\)
----
\(q < 0\)
----
\(\sin (-\theta) = - \sin \theta\)
----
\(y = a \cos (k \theta + p) + q\)
----
\(k > 1\)
----
\(k < -1\)
----
\(0  <k <1\)
----
\(-1  <k <0\)
----
\(p > 0\)
----
\(p < 0\)
----
\(q > 0\)
----
\(q < 0\)
----
\(\cos (-\theta) = \cos \theta\)
----
\(y = a \tan (k \theta + p) + q\)
----
\(k > 1\)
----
\(k < -1\)
----
\(0  <k <1\)
----
\(-1  <k <0\)
----
\(p > 0\)
----
\(p < 0\)
----
\(q > 0\)
----
\(q < 0\)
----
\(\tan (-\theta) = - \tan \theta\)
----
\(n \in \mathbb{Z}\)
----
\(a<0\)
----
\(f(x)=a{\left(x+p\right)}^{2}+q\)
----
\(\left\{f(x):f(x)\in \left(-\infty ,q\right]\right\}\)
----
\[\begin{align}
(x + p)^2 \geq 0\\
a(x + p)^2   \leq 0 \\
a(x + p)^2 + q   < q \\
\therefore f(x)   < q  
\end{align}\]
----
\(\left(2;7\right)\)
----
\(f(x)=-2{x}^{2}-4ax+k\)
----
\(a\)
----
\(k\)
----
\(a = -2; k= -1\)
----
\(f(x)=a{x}^{2}+bx\)
----
\(\left(3;9\right)\)
----
\(a=-1\)
----
\(b=6\)
----
\[\begin{align}
f(x) & = ax^2 + bx \\
 x &= -{b}{2a} \\
 \therefore 3 &= -\frac{b}{2a}\\
6a &= -b\\
 \text{subst. } (3;9) \quad 9 &=  a(3)^2 + b(3) \\
   9 &=  9a  + 3b  \\
   9 &=  9a  + 3(-6a)  \\
   9 &=  9a  -18a  \\
   9 &=  -9a  \\
   \therefore -1 &=  a  \\
   \therefore b &=  6   
\end{align}\]
----
\(f(x)={x}^{2}-2x+3\)
----
\(f\)
----
\(y={x + 2}^{2} + 2\)
----
\(x\)
----
\(y={x - 1}^{2} + 5\)
----
\(\left(-1;-4\right)\)
----
\((-1;0)\)
----
\(y=\frac{2}{x}\)
----
\(-4\le x\le 4\)
----
\(y=\frac{2}{x - 3} - 1\)
----
\(y=\frac{k}{(x+p)} + q\)
----
\(y=2\)
----
\(x=1\)
----
\(\left(2;3\right)\)
----
\(y=\frac{1}{(x-1)} + 2\)
----
\(y\)
----
\(y={a}^{x}\)
----
\(y={a}^{x+1}\)
----
\(y={a}^{x}+1\)
----
\(x\)
----
\(y={a}^{x} + 1\)
----
\(y={a}^{x}\)
----
\(y={a}^{x + 1}\)
----
\(f(x) = 1 + a \dottimes 2^x\)
----
\(a\)
----
\(a\)
----
\[\begin{align}
   0 &=  a \times 2^0 + 1 \\
   \therefore -1 &=  a  
\end{align}\]
----
\(a = -1\)
----
\(f(-15)\)
----
\(x\)
----
\(f\)
----
\(f\)
----
\(h\)
----
\(h\)
----
\(h(x) = -2^{(x -2)} + 1\)
----
\(f(x) = a \dottimes b^x\)
----
\(a ≠ 0\)
----
\(f\)
----
\(a\)
----
\(a = 256\)
----
\(f\)
----
\(f(x) = 256 \left( \frac{3}{4} \right)^x\)
----
\(f(13)\)
----
\(f\)
----
\(h\)
----
\(h(x)=f(-x)\)
----
\(p\)
----
\(k\)
----
\(y = \sin 3\theta\)
----
\(y = -\cos 2\theta\)
----
\(y = \tan \frac{1}{2}\theta\)
----
\(y = 2\sin 2\theta\)
----
\((270 \pm x)\)
----
\(ABC\)
----
\(P(x;y)\)
----
\(Q(-x;y)\)
----
\(x\)
----
\((OP \text{ and } OQ)\)
----
\(P(x;y)\)
----
\(P\hat{O}X = \alpha\)
----
\(Q(-x;y)\)
----
\(Q\hat{O}X = \beta\)
----
\[\begin{align}
    OP^2 &= x^2 + y^2   \\
    \text{And }OQ^2 &=  (-x)^2 + y^2 \\
    &= x^2 + y^2   \\
    \therefore OP &= OQ
    \end{align}\]
----
\(OP = OQ = r\)
----
\[\begin{align}
    \sin \alpha &=  \frac{y}{r} \\
    \cos \alpha &=  \frac{x}{r}\\
    \tan \alpha &=  \frac{y}{x}
    \end{align}\]
----
\(-x < 0\)
----
\[\begin{align}
    \sin \beta &=   \frac{y}{r} \\
    \cos \beta &=  -\frac{x}{r}\\
    \tan \beta &=  -\frac{y}{x}
    \end{align}\]
----
\(x\)
----
\(y\)
----
\(\cos θ\)
----
\(\frac{\sqrt{3}}{2}\)
----
\(\frac{1}{\sqrt{2}}\)
----
\(\frac{1}{2}\)
----
\(\sin θ\)
----
\(\frac{1}{2}\)
----
\(\frac{1}{\sqrt{2}}\)
----
\(\frac{\sqrt{3}}{2}\)
----
\(\tan θ\)
----
\(\frac{1}{\sqrt{3}}\)
----
\(\sqrt{3}\)
----
\(a\)
----
\(b\)
----
\(TUW\)
----
\(a\)
----
\(b\)
----
\(b\)
----
\(\theta\)
----
\(MNP\)
----
\(\theta\)
----
\(\theta\)
----
\(\theta\)
----
\(\theta\)
----
\(\sin p + 3 \sin p = 4 \sin p\)
----
\(\dfrac{\sin q}{\cos q} = \tan q\)
----
\(\cos (p - q) = \cos p - \cos q\)
----
\(\sin (2p) = 2\sin p \cos p\)
----
\(\dfrac{\sin \theta}{\cos \theta} = 4\frac{2}{3}\)
----
\(A\)
----
\(BC\)
----
\(T\)
----
\(TC\)
----
\(AT\)
----
\(B\hat{A}T\)
----
\(2x\)
----
\(2y\)
----
\(\beta\)
----
\(\cos \beta\)
----
\(\cos^2 \beta + \sin^2 \beta\)
----
\(\sin \theta\)
----
\(\cos \theta\)
----
\(\dfrac{\sin \theta}{\cos \theta}\)
----
\(\tan \theta\)
----
\(\theta\)
----
\(\tan \theta\)
----
\[\tan \theta = \frac{\text{opposite side}}{\text{adjacent side}}\]
----
\(x, y\)
----
\(r\)
----
\[\begin{align}
    \tan \theta &= \frac{y}{x}  \\
    &= \frac{y}{x} \times \frac{r}{r} \\
    &= \frac{y}{r} \times \frac{r}{x} \\
    &= \frac{y}{r} \div \frac{x}{r} \\
    &= \sin \theta \div \cos \theta \\
    &= \frac{\sin \theta}{\cos \theta}
    \end{align}\]
----
\[\tan \theta = \frac{\sin \theta}{\cos \theta}\]
----
\(\tan \theta\)
----
\(\cos \theta = 0\)
----
\(k\)
----
\(x, y\)
----
\(r\)
----
\(x, y\)
----
\(r\)
----
\[\begin{align}
    \sin^2 \theta + \cos^2 \theta &= \left( \frac{y}{r} \right)^2 + \left( \frac{x}{r} \right)^2 \\
    &= \frac{y^2}{r^2} + \frac{x^2}{r^2} \\
    &= \frac{y^2 + x^2}{r^2} \\
    &= \frac{r^2}{r^2} \\
    &= 1
    \end{align}\]
----
\[\sin^2 \theta + \cos^2 \theta = 1\]
----
\(\sin^2 \theta = 1 - \ldots\ldots\)
----
\(\cos \theta = \pm \sqrt{\ldots\ldots}\)
----
\(\sin^2 \theta = (1 + \ldots\ldots)(1 - \ldots\ldots)\)
----
\(\cos^2 \theta - 1 = \ldots\ldots\)
----
\(\sin^2 \theta + \cos^2 \theta\)
----
\(\dfrac{1 + \sin \theta}{1 + \sin \theta}\)
----
\(\tan^2 \theta \times \cos^2 \theta\)
----
\(\dfrac{1}{\cos^2 \theta} - \tan^2 \theta\)
----
\[\begin{align}
    \tan^2 \theta \times \cos^2 \theta &= \left( \frac{\sin \theta}{\cos \theta} \right)^2 \times \cos^2 \theta \\
    &= \frac{\sin^2 \theta}{\cos^2 \theta} \times \cos^2 \theta \\
    &= \sin^2 \theta
    \end{align}\]
----
\[\begin{align}
    \frac{1}{\cos^2 \theta} - \tan^2 \theta &= \frac{1}{\cos^2 \theta} - \left( \frac{\sin \theta}{\cos \theta} \right)^2 \\
    &= \frac{1}{\cos^2 \theta} - \frac{\sin^2 \theta}{\cos^2 \theta} \\
    &= \frac{1 - \sin^2 \theta}{\cos^2 \theta} \\
    &= \frac{\cos^2 \theta}{\cos^2 \theta} \\
    &= 1
    \end{align}\]
----
\(\dfrac{1 - \sin \alpha}{\cos \alpha} = \dfrac{\cos \alpha}{1 + \sin \alpha}\)
----
\(\cos \theta \ne 0\)
----
\(\sin \theta + 1 \ne 0\)
----
\[\begin{align}
                                \text{LHS} &= \frac{1 - \sin \alpha}{\cos \alpha}  \\
                                &= \frac{1 - \sin \alpha}{\cos \alpha} \times \frac{1 + \sin \alpha}{1 + \sin \alpha}
                                \end{align}\]
----
\[\begin{align}
                                \text{LHS} &= \frac{1 - \sin^2 \alpha}{\cos \alpha(1 + \sin \alpha)} \\
                                &= \frac{\cos^2 \alpha}{\cos \alpha(1 + \sin \alpha)} \\
                                &= \frac{\cos \alpha}{1 + \sin \alpha} \\
                                &= \text{RHS}
                                \end{align}\]
----
\(\dfrac{\sin\alpha}{\tan\alpha}\)
----
\(\cos^2 \theta \tan^2 \theta + \tan^2 \theta \sin^2 \theta\)
----
\(1-\sin \theta \cos \theta \tan \theta\)
----
\(\left( \dfrac{1 - \cos^2\beta}{\cos^2 \beta}\right) - \tan^2 \beta\)
----
\(\dfrac{1+\sin\theta}{\cos\theta} = \dfrac{\cos\theta}{1-\sin\theta}\)
----
\[\begin{aligned} 
\text{LHS } &= \dfrac{1+\sin \theta }{\cos\theta }\\
&=\dfrac{(1+\sin\theta)}{\cos\theta})\times \dfrac{(1-\sin\theta)}{1-\sin\theta}\\
&=\dfrac{1-\sin^2\theta}{\cos\theta(1-\sin\theta)} \\
&=\dfrac{\cos^2\theta }{\cos\theta(1-\sin\theta)} \\
&=\dfrac{\cos\theta}{(1-\sin\theta)} \\
&= \text{RHS} 
\end{aligned}\]
----
\(\tan \theta\)
----
\({\sin}^{2}\alpha + \left(\cos\alpha - \tan\alpha \right)\left(\cos\alpha +\tan\alpha \right)=1-{\tan}^{2}\alpha\)
----
\[\begin{aligned} 
\text{LHS } &= \sin^2 \alpha + (\cos\alpha-\tan\alpha) (\cos\alpha+\tan\alpha) \\
&=\sin^2\alpha+\cos^2\alpha-\tan^2\alpha \\
&=1 - \tan^2\alpha \\
&= \text{RHS} 
\end{aligned}\]
----
\(\tan \theta\)
----
\(\dfrac{1}{\cos\theta}-\dfrac{\cos\theta {\tan}^{2}\theta }{1}=\cos\theta\)
----
\[\begin{aligned} 
\text{LHS } &= \dfrac{1}{\cos\theta}-\dfrac{\cos\theta \tan^2\theta}{1}\\
&=\dfrac{1-\cos^2\theta \times \tan^2\theta}{\cos\theta}\\
&=\dfrac{1-\cos^2\theta \times \dfrac{\sin^2\theta}{\cos^2\theta}}{\cos\theta}\\
&=\dfrac{1-\sin^2\theta}{\cos\theta}\\
&=\dfrac{\cos^2\theta}{\cos\theta}\\
&=\cos\theta\\
&= \text{RHS} 
\end{aligned}\]
----
\(\tan \theta\)
----
\(\dfrac{2\sin\theta\cos\theta}{\sin\theta +\cos\theta }=\sin\theta +\cos\theta -\dfrac{1}{\sin\theta +\cos\theta }\)
----
\[\begin{aligned} 
\text{RHS } &= \sin\theta+\cos\theta-\dfrac{1}{\sin\theta+\cos\theta} \\
&=\dfrac{\sin^2\theta+\sin\theta \cos\theta+\cos\theta \sin\theta+\cos^2\theta-1}{\sin\theta+\cos\theta} \\
&=\dfrac{1+2\sin\theta \cos\theta-1}{\sin\theta+\cos\theta} \\
&=\dfrac{2\sin\theta \cos\theta}{\sin\theta+\cos\theta} \\
&= \text{LHS} 
\end{aligned}\]
----
\(\left(\dfrac{\cos\beta}{\sin\beta }+\tan\beta \right) \cos\beta =\dfrac{1}{\sin\beta }\)
----
\[\begin{aligned} 
\text{ LHS} &=  \left ( \dfrac{\cos\beta}{\sin\beta}+\dfrac{\sin\beta}{\cos\beta} \right ) \\
&=\left ( \dfrac{\cos^2\beta + \sin^2\beta}{\sin\beta\cos\beta} \right )\cos\beta\\
&=\dfrac{1}{\sin\beta}\\
&= \text{RHS } 
\end{aligned}\]
----
\(\dfrac{1}{1 + \sin \theta} + \dfrac{1}{1 - \sin \theta} = d\frac{2 \tan \theta}{\sin \theta \cos \theta}\)
----
\[\begin{aligned} 
\text{ LHS} &= \dfrac{1-\sin\theta+1+\sin\theta}{(1+\sin\theta)(1-\sin\theta)} \\
&=\dfrac{2}{1-\sin^2\theta} \\
&=\dfrac{2}{\cos^2\theta}
\end{aligned}\]
----
\[\begin{aligned} 
\text{ RHS} &= \dfrac{2 \tan \theta}{\sin \theta \cos \theta} \\
&=\dfrac{2 \sin \theta}{\sin \theta \cos \theta \cos \theta} \\
&=\dfrac{2}{\cos^2\theta} \\
\therefore \text{LHS } &= \text{RHS } 
\end{aligned}\]
----
\(\dfrac{(1 + \tan^2 \alpha) \cos \alpha}{(1 - \tan \alpha)} = \dfrac{1}{\cos \alpha - \sin \alpha}\)
----
\[\begin{aligned} 
\text{ LHS} &= \dfrac{(1 + \tan^2 \alpha) \cos \alpha}{(1 - \tan \alpha)} \\
&= \dfrac{\left (1+\frac{\sin^2\alpha}{\cos^2\alpha}  \right )\cos\alpha}{\left ( 1-\frac{\sin\alpha}{\cos\alpha} \right )} \\
&=\dfrac{\left ( \frac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha} \right )\cos\alpha}{\frac{\cos\alpha-\sin\alpha}{\cos\alpha}}\\
&=\dfrac{1}{\cos\alpha}\times \frac{\cos\alpha}{\cos\alpha-\sin\alpha} \\
&=\dfrac{1}{\cos\alpha-\sin\alpha} \\
&= \text{RHS } 
\end{aligned}\]
----
\(\theta\)
----
\(P(\sqrt{3};1)\)
----
\(P'\)
----
\(OP\)
----
\(x\)
----
\(P\)
----
\(P'\)
----
\(y\)
----
\(P'\)
----
\(\sin \theta\)
----
\(\cos \theta\)
----
\(\tan \theta\)
----
\(P'\)
----
\(\theta\)
----
\(P(\sqrt{3};1)\)
----
\(P'\)
----
\(OP\)
----
\(x\)
----
\(P\)
----
\(P'\)
----
\(x\)
----
\(y\)
----
\(P'\)
----
\(P'\)
----
\(\theta\)
----
\(\theta\)
----
\(t\)
----
\((- \theta)\)
----
\((- \theta)\)
----
\(x\)
----
\(\theta\)
----
\(-\theta\)
----
\(P(\sqrt{3};1)\)
----
\(P'\)
----
\(OP\)
----
\(x\)
----
\(P\)
----
\(P'\)
----
\(x\)
----
\(y = 0\)
----
\(P'\)
----
\(P'\)
----
\(P'\)
----
\(\sin (-\theta)\)
----
\(\cos (-\theta)\)
----
\(\tan (-\theta)\)
----
\(- \theta\)
----
\(\sin (- \theta) = \ldots \ldots\)
----
\(\cos (- \theta) = \ldots \ldots\)
----
\(\tan (- \theta) = \ldots \ldots\)
----
\(\theta\)
----
\(y = \sin \theta\)
----
\(\sin \theta\)
----
\(k\)
----
\(f\)
----
\(\cos \beta\)
----
\(\sin (-\alpha) = - \sin \alpha\)
----
\[\begin{aligned} 
\sin(-\alpha) &=\frac{-y}{r} \qquad \text{(lies in III quad)}\\
&=-\sin\alpha 
 \end{aligned}\]
----
\(\cos (-\alpha) =  -\cos \alpha\)
----
\[\begin{aligned} 
\cos(-\alpha)&=\frac{-x}{r} \qquad \text{(lies in III quad)}\\
&=-\cos\alpha
\end{aligned}\]
----
\(t\)
----
\(\triangle ABC\)
----
\[\begin{align}
  \sin \hat{C} &= \dfrac{c}{b} = \cos \ldots \\
  \cos \hat{C} &= \dfrac{a}{b} = \sin \ldots
  \end{align}\]
----
\(P(\sqrt{3};1)\)
----
\(P'\)
----
\(OP\)
----
\(x\)
----
\(P\)
----
\(P'\)
----
\(y = x\)
----
\(P'\)
----
\(P'\)
----
\(\theta\)
----
\(P(\sqrt{3};1)\)
----
\(P'\)
----
\(OP\)
----
\(x\)
----
\(P\)
----
\(P'\)
----
\(P'\)
----
\(P'\)
----
\(\theta\)
----
\(f\left(A\right)=g\left(B\right)\)
----
\(A\)
----
\(B\)
----
\(t\)
----
\(t = \sin \theta\)
----
\(p\)
----
\(\theta\)
----
\(\theta\)
----
\(-\theta\)
----
\((\theta)\)
----
\(A\)
----
\(B\)
----
\(A + B = \ldots\)
----
\(\frac{A}{B} = \ldots\)
----
\(\theta\)
----
\(\tan \theta = 5\)
----
\(\theta\)
----
\(\theta\)
----
\(\tan \theta\)
----
\(\tan \theta\)
----
\(\theta\)
----
\(\alpha\)
----
\(\cos \alpha\)
----
\(\cos \alpha\)
----
\(\alpha\)
----
\(\beta\)
----
\(\sin \beta\)
----
\(\sin \beta\)
----
\(\beta\)
----
\(\alpha\)
----
\(4 \cos \alpha = 2\)
----
\(\tan \alpha = 5\frac{1}{4}\)
----
\(5 \sin \alpha = 3\)
----
\(\theta\)
----
\(\cos \theta + \frac{3}{4} = 0\)
----
\(3 \tan \theta = 20\)
----
\(2 \cos \theta = \frac{4}{5}\)
----
\(\sin \theta\)
----
\(\sin \theta\)
----
\(k \in \mathbb{Z}\)
----
\(k\)
----
\(k = 4\)
----
\(k = -2\)
----
\(\cos \theta\)
----
\(\cos \theta\)
----
\(k \in \mathbb{Z}\)
----
\(\theta\)
----
\(k\)
----
\(k = 2\)
----
\(k = -5\)
----
\(\tan x\)
----
\(k \in \mathbb{Z}\)
----
\(\alpha\)
----
\(k\)
----
\(\alpha\)
----
\(k = 0\)
----
\(k = 1\)
----
\(k = 2\)
----
\(k = -1\)
----
\(k = -2\)
----
\(k = -3\)
----
\(k = -4\)
----
\(k \in \mathbb{Z}\)
----
\(k \in \mathbb{Z}\)
----
\(\cos \alpha = 1\)
----
\(4 \sin \theta = 0\)
----
\(\cos \theta = -1\)
----
\(4 \cos \theta +3 = 1\)
----
\(\sin 2\theta  = -\frac{\sqrt{3}}{2}\)
----
\(\sin 3\alpha = -1\)
----
\(2 \sin \frac{3\theta}{2} = -1\)
----
\(4 \sin^2 \theta = 3\)
----
\(\sin \theta\)
----
\(k \in \mathbb{Z}\)
----
\(\theta\)
----
\(2 \cos^2 \theta - \cos \theta - 1 = 0\)
----
\[\begin{align}
                                2 \cos^2 \theta - \cos \theta - 1 &= 0 \\
                                 (2 \cos \theta + 1)(\cos \theta - 1) &= 0 \\
                                 \therefore 2 \cos \theta + 1 = 0  &\text{ or } \cos \theta - 1 = 0
                                \end{align}\]
----
\(\theta\)
----
\(k \in \mathbb{Z}\)
----
\(k\)
----
\(\theta\)
----
\(k\)
----
\(k = -1\)
----
\(k = 0\)
----
\(k = 1\)
----
\(y = \cos \theta\)
----
\[\begin{align}
            2y^2  - y - 1 &= 0 \\
            (2 y + 1)(y - 1) &= 0 \\
            \therefore y = -\frac{1}{2} &\text{ or } y = 1
          \end{align}\]
----
\(y = \cos \theta\)
----
\(\theta\)
----
\(\alpha\)
----
\(2 \sin^2 \alpha - \sin \alpha \cos \alpha = 0\)
----
\[\begin{align}
                                2 \sin^2 \alpha - \sin \alpha \cos \alpha &= 0 \\
                                 \sin \alpha (2 \sin \alpha  - \cos \alpha) &= 0 \\
                                 \therefore  \sin \alpha  = 0  &\text{ or } 2 \sin \alpha  - \cos \alpha = 0
                                \end{align}\]
----
\(\alpha\)
----
\[\begin{align}
          2 \sin \alpha  - \cos \alpha  &= 0 \\
          2 \sin \alpha  &= \cos \alpha
          \end{align}\]
----
\(\cos \alpha\)
----
\(k \in \mathbb{Z}\)
----
\(k\)
----
\(\alpha\)
----
\(k\)
----
\(k = 0\)
----
\(k = 1\)
----
\(k = 2\)
----
\(\cos 2\theta = 0\)
----
\(2 \cos \frac{\theta}{2} - \sqrt{3} = 0\)
----
\(\sin 2 \beta = \cos 2 \beta\)
----
\(\theta\)
----
\(\sin^2 \theta + \frac{1}{2} \sin \theta = 0\)
----
\(2 \cos^2 \theta - 3 \cos \theta = 2\)
----
\(3 \tan^2 \theta + 2 \tan \theta = 0\)
----
\(\sin^2 \theta - 4\cos^2 \theta = 0\)
----
\(\beta\)
----
\(\triangle ABC\)
----
\(\triangle ABC\)
----
\(\frac{1}{2} \times \ldots \times AC\)
----
\(\sin \hat{B} = \ldots\)
----
\(AC = \ldots \times \ldots\)
----
\(\triangle ABC\)
----
\(\ldots \times \ldots \times \ldots \times \ldots\)
----
\(\triangle A'B'C'\)
----
\(\triangle A'B'C'\)
----
\(\triangle ABC\)
----
\(\triangle A'B'C'\)
----
\(\triangle PQR\)
----
\(\triangle ABC\)
----
\(AB = c, BC = a\)
----
\(AC = b\)
----
\(h\)
----
\(A\)
----
\(BC\)
----
\(\triangle ABC\)
----
\[\begin{align}
          \sin \hat{B}&= \frac{h}{c} \\
          \therefore h &= c \sin \hat{B}
           \end{align}\]
----
\[\begin{align}
          \text{Area } \triangle ABC &= \frac{1}{2} \times a \times h \\
            &= \frac{1}{2} \times a \times c \sin \hat{B} \\
            \therefore \text{Area } \triangle ABC  &= \frac{1}{2} ac \sin \hat{B}
           \end{align}\]
----
\[\begin{align}
          \sin \hat{C}&= \frac{h}{b} \\
          \therefore h &= b \sin \hat{C}
           \end{align}\]
----
\[\begin{align}
          \text{Area } \triangle ABC &= \frac{1}{2} \times a \times h \\
            &= \frac{1}{2} ab \sin \hat{C}
           \end{align}\]
----
\(B\)
----
\(AC\)
----
\(\triangle ABC = \frac{1}{2} bc \sin \hat{A}\)
----
\(\triangle ABC\)
----
\[\begin{align}
        \text{Area } \triangle ABC
        &= \frac{1}{2} bc \sin \hat{A} \\
        &= \frac{1}{2} ac \sin \hat{B} \\
        &= \frac{1}{2} ab \sin \hat{C}
        \end{align}\]
----
\(⃤ABC\)
----
\(\triangle ABC\)
----
\(a\)
----
\(\triangle ABC\)
----
\(\triangle DEF = \frac{1}{2} df \sin \hat{E}\)
----
\(h\)
----
\(DH\)
----
\(DH \perp EF\)
----
\(DH = h\)
----
\(D\hat{E}F = \hat{E}_1\)
----
\(D\hat{E}H = \hat{E}_2\)
----
\(\triangle DHE\)
----
\(\triangle DEF\)
----
\(\triangle DEF\)
----
\[\begin{align}
            \text{Area } &= \frac{1}{2} d \times h \\
            &= \frac{1}{2} df \sin \hat{E}_1
          \end{align}\]
----
\(\triangle PQR\)
----
\[\begin{align}
        \text{Area } \triangle PQR &= \frac{1}{2} qr \sin \hat{P} \\
        &= \frac{1}{2} pr \sin \hat{Q} \\
        &= \frac{1}{2} pq \sin \hat{R}
        \end{align}\]
----
\(⃤PQR\)
----
\(r=10\)
----
\(p=7\)
----
\(p=8\)
----
\(q=9\)
----
\(\triangle XYZ\)
----
\(⃤ABC\)
----
\(\hat{C}\)
----
\(\triangle ABC\)
----
\(AC = 15\)
----
\(BC = 11\)
----
\(\hat{B}\)
----
\(\triangle ABC\)
----
\(CF \perp AB\)
----
\(\triangle CBF\)
----
\[\begin{align}
                    \frac{CF}{\ldots} &= \sin \hat{B}   \\
                    \therefore CF &= \ldots \times \sin \hat{B}
                    \end{align}\]
----
\(\triangle CAF\)
----
\[\begin{align}
                    \frac{CF}{15} &= \ldots    \\
                    \therefore CF &= 15 \times \ldots
                    \end{align}\]
----
\[\begin{align}
                    CF &=  15 \times \ldots   \\
                    \text{and } CF &= \ldots \times \sin \hat{B} \\
                    \therefore  15 \times \ldots &= \ldots \times \sin \hat{B}  \\
                    \therefore \sin \hat{B} &= \ldots \ldots \ldots \\
                    \therefore \hat{B} &= \ldots
                    \end{align}\]
----
\(\triangle ABC\)
----
\[\begin{align}
                    \text{Area } \triangle ABC &= \frac{1}{2} AB \times AC \times \ldots  \\
                            &= \frac{1}{2} AB \times \ldots  \times \ldots
                    \end{align}\]
----
\[\text{Area } \triangle ABC = \frac{1}{2} AB \times \ldots \times \sin \hat{B}\]
----
\(\hat{B}\)
----
\[\begin{align}
                    \frac{1}{2} AB \times \ldots \times \sin \hat{B} &= \frac{1}{2} AB \times \ldots  \times \ldots \\
                      \therefore   \ldots \times \sin \hat{B}   &= \ldots  \times \ldots \\
                      \therefore  \sin \hat{B}   &= \ldots  \times \ldots \\
                                        \therefore \hat{B} &= \ldots
                    \end{align}\]
----
\(\triangle PQR\)
----
\(ABC\)
----
\(AB = c, BC = a\)
----
\(AC = b\)
----
\(h\)
----
\(F\)
----
\(\triangle ABF\)
----
\[\begin{align}
          \sin \hat{B}&= \frac{h}{c} \\
          \therefore h &= c \sin \hat{B}
           \end{align}\]
----
\(\triangle ACF\)
----
\[\begin{align}
          \sin \hat{C}&= \frac{h}{b} \\
          \therefore h &= b \sin \hat{C}
           \end{align}\]
----
\[\begin{align}
          c \sin \hat{B}&= b \sin \hat{C} \\
          \therefore \frac{\sin \hat{B}}{b} &= \frac{\sin \hat{C}}{c} \\
          \text{or } \frac{b}{\sin \hat{B}} &= \frac{c}{\sin \hat{C}}
           \end{align}\]
----
\(B\)
----
\(AC\)
----
\[\begin{align}
           \frac{\sin \hat{A}}{a} &= \frac{\sin \hat{C}}{c} \\
          \text{or } \frac{a}{\sin \hat{A}} &= \frac{c}{\sin \hat{C}}
           \end{align}\]
----
\(B\)
----
\(AC\)
----
\(\triangle ABC = \frac{1}{2} bc \sin \hat{A}\)
----
\(\triangle ABC\)
----
\[\begin{align}
          \text{Area } \triangle ABC &= \frac{1}{2} ac \sin \hat{B} \\
                                                                   &= \frac{1}{2} ab \sin \hat{C} \\
          \therefore \frac{1}{2} ac \sin \hat{B} &= \frac{1}{2} ab \sin \hat{C} \\
             c \sin \hat{B} &=  b \sin \hat{C} \\
           \frac{\sin \hat{B}}{b} &= \frac{\sin \hat{C}}{c} \\
          \text{or } \frac{b}{\sin \hat{B}} &= \frac{c}{\sin \hat{C}}
           \end{align}\]
----
\(\triangle ABC\)
----
\[\begin{align*}
          \dfrac{\sin \hat{A}}{a} &= \dfrac{\sin \hat{B}}{b} = \dfrac{\sin \hat{C}}{c} \\
          \dfrac{a}{\sin \hat{A}} &= \dfrac{b}{\sin \hat{B}} = \dfrac{c}{\sin \hat{C}}
        \end{align*}\]
----
\(\triangle TRS\)
----
\(TR = 30\)
----
\(RS, ST\)
----
\(T\hat{R}S\)
----
\(RS = t\)
----
\(ST = r\)
----
\(TR = s\)
----
\(T\hat{R}S\)
----
\(t\)
----
\(r\)
----
\(\triangle MNP\)
----
\(MS \perp NP\)
----
\(\triangle MSN\)
----
\(\triangle MSP\)
----
\[\begin{align}
          \sin \hat{P}&= \frac{h}{n} \\
          \therefore h &= n \sin \hat{P}
          \end{align}\]
----
\[\begin{align}
            p \sin \hat{N}_1  &= n \sin \hat{P} \\
            \therefore \frac{\sin \hat{N}_1}{n}  &= \frac{\sin \hat{P}}{p}  \\
            \text{or }  \frac{n}{\sin \hat{N}_1}  &= \frac{p}{\sin \hat{P}}
          \end{align}\]
----
\(\triangle NMP\)
----
\(\triangle NMP'\)
----
\(\triangle ABC\)
----
\(AB = 82\)
----
\(BC = 65\)
----
\(\triangle ABC\)
----
\(\hat{C}\)
----
\(\triangle ABC\)
----
\(BC\)
----
\(\hat{A}\)
----
\(AB\)
----
\(\triangle ABC\)
----
\(\triangle A'B'C'\)
----
\(AC\)
----
\(BC\)
----
\(⃤PQR\)
----
\(r=3\)
----
\(⃤KLM\)
----
\(m=1\)
----
\(\triangle ABC\)
----
\(⃤XYZ\)
----
\(x=50\)
----
\(⃤ABC\)
----
\(AB\)
----
\(BC\)
----
\(⃤RST\)
----
\(ST\)
----
\(\triangle KMS\)
----
\(m\)
----
\(\triangle ABD\)
----
\(\triangle BCD\)
----
\(BC\)
----
\(\triangle ABC\)
----
\(BC\)
----
\(\triangle ABC, AB = 21, AC = 17\)
----
\(\hat{B}\)
----
\(CB\)
----
\(CH \perp AB\)
----
\(AH = c\)
----
\(HB = \ldots\)
----
\(\triangle CHB\)
----
\[\begin{align}
                    CB^2 &= BH^2 + CH^2   \\
                     &= (\ldots)^2 + CH^2 \\
                     &= 21^2 -(2)(21)c + c^2 + CH^2 \ldots \ldots (1)
                    \end{align}\]
----
\(\triangle CHA\)
----
\[\begin{align}
                    CA^2 &= c^2 + CH^2   \\
                     17^2 &= c^2 + CH^2 \ldots \ldots (2)
                    \end{align}\]
----
\((2)\)
----
\((1)\)
----
\[CB^2 = 21^2 -(2)(21)c + 17^2\]
----
\(c\)
----
\(\triangle CHA\)
----
\(\triangle PQR\)
----
\(\triangle ABC\)
----
\(CD \perp AB\)
----
\(\triangle DCB\)
----
\(a^2 = (c-d)^2 + h^2\)
----
\(\triangle ACD\)
----
\(b^2 = d^2 + h^2\)
----
\(h^2\)
----
\[\begin{align*}
    a^2 &= (c-d)^2 + h^2 \\
\therefore h^2 &= a^2 - (c-d)^2 \\
    \text{And } b^2 &= d^2 + h^2 \\
\therefore h^2 &= b^2  - d^2 \\
  \therefore b^2 - d^2 &= a^2 - (c-d)^2 \\
  a^2 &= b^2 + (c^2 - 2cd + d^2) - d^2 \\
  &= b^2 + c^2 - 2cd
\end{align*}\]
----
\(d\)
----
\(⃤ACD\)
----
\(\cos\hat{A} = \frac{d}{b}\)
----
\(d = b \cos \hat{A}\)
----
\({a}^{2}={b}^{2}+{c}^{2}-2bc \cos \hat{A}\)
----
\(\triangle ABC\)
----
\[\begin{align*}
          a^2 &= b^2 + c^2 - 2bc \cos \hat{A} \\
          b^2 &= a^2 + c^2 - 2ac \cos \hat{B} \\
          c^2 &= a^2 + b^2 - 2ab \cos \hat{C}
        \end{align*}\]
----
\(QR\)
----
\(\hat{A}\)
----
\(⃤ABC\)
----
\(b=4\)
----
\(c=9\)
----
\(⃤RST\)
----
\(RS=14\)
----
\(ST=26\)
----
\(RT=16\)
----
\(⃤KLM\)
----
\(KL=5\)
----
\(LM=10\)
----
\(KM=7\)
----
\(⃤JHK\)
----
\(JH=13\)
----
\(HK=8\)
----
\(⃤DEF\)
----
\(d=4\)
----
\(e=5\)
----
\(f=7\)
----
\(⃤XYZ\)
----
\(⃤JHK\)
----
\(JH=6\)
----
\(HK=4\)
----
\(JK=3\)
----
\(⃤PQR\)
----
\(p=50\)
----
\(q=70\)
----
\(r=60\)
----
\(Q\)
----
\(P\)
----
\(R\)
----
\(QR\)
----
\(PQ\)
----
\(WXYZ\)
----
\(WX\parallel YZ\)
----
\(XZ\)
----
\(XY\)
----
\(ABCD\)
----
\(AB\parallel CD\)
----
\(AB=x\)
----
\(B\hat{A}D=a\)
----
\(B\hat{C}D=b\)
----
\(B\hat{D}C=c\)
----
\(CD\)
----
\(x\)
----
\(a\)
----
\(b\)
----
\(c\)
----
\(DC = \frac{x \sin a \sin (b+c)}{\sin(a+c) \sin b}\)
----
\(X\)
----
\(Z\)
----
\(Y\)
----
\(X\)
----
\(Z\)
----
\(X\hat{Y}Z\)
----
\(XY=x\)
----
\(X\hat{Y}Z=\theta\)
----
\(XZ = x\sqrt{2(1- \cos \theta)}\)
----
\[\begin{align}
  XZ^2 &= 2x^2   - 2x^2 \cos \theta \\
   &= 2x^2 \left(1   -  \cos \theta \right) \\
   \therefore XZ &= \sqrt{2x^2 \left(1   -  \cos \theta \right)} \\
                            &= x\sqrt{2 \left(1   -  \cos \theta \right)}
\end{align}\]
----
\(XZ\)
----
\(WXYZ\)
----
\(x\)
----
\(\alpha\)
----
\(\beta\)
----
\(\theta\)
----
\(\phi\)
----
\[\begin{align}
  \text{In } \triangle ABE, \quad \frac{x}{EB}    &= \cos \theta \\
\therefore EB    &= \frac{x}{\cos \theta} \\
  \text{In } \triangle BDC, \quad \frac{DB}{\sin \alpha}    &= \frac{x}{\sin \beta} \\
\therefore DB     &= \frac{x \sin \alpha }{\sin \beta} \\
                \therefore \text{Area} \triangle BED &= \frac{1}{2}(BE)(BD)\sin (\beta - \theta)\\
                    &= \frac{1}{2} \times \frac{x}{\cos \theta} \times \frac{x \sin \alpha }{\sin \beta} \times \sin (\beta - \theta)\\
            &=  \frac{x^2 \sin \alpha \sin (\beta - \theta) }{2 \cos \theta \sin \beta}
\end{align}\]
----
\(DC = \frac{x \sin \alpha }{ \sin \beta }\)
----
\(\cos^2\theta + \sin^2\theta = 1\)
----
\(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\)
----
\(\sin (-\theta) = - \sin \theta\)
----
\(\cos (-\theta) = \cos \theta\)
----
\(\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\)
----
\(\triangle ABC = \frac{1}{2} bc \sin A\)
----
\(a^2 = b^2 + c^2 - 2 bc \cos A\)
----
\(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)
----
\(\triangle ABC = \frac{1}{2} ac \sin B\)
----
\(b^2 = a^2 + c^2 - 2 ac \cos B\)
----
\(\triangle ABC = \frac{1}{2} ab \sin C\)
----
\(c^2 = a^2 + b^2 - 2 ab \cos C\)
----
\(k \in \mathbb{Z}\)
----
\(\sin^2A\)
----
\(1\frac{1}{4}\)
----
\(\cos \alpha\)
----
\(\dfrac{1}{(\cos x - 1)(\cos x + 1)} = \dfrac{-1}{\tan^2x \cos^2x}\)
----
\[\begin{align}
\text{ RHS  }&=\frac{-1}{\frac{\sin^2x}{\cos^2x}\dottimes\frac{\cos^2x}{1}} \\
&=\frac{-1}{\sin^2x} \\
\text{ LHS  } &=\frac{1}{\cos^2x-1} \\
&=\frac{1}{-(1-\cos^2x)} \\
&=-\frac{1}{\sin^2x} \\
&=\text{ RHS  }
\end{align}\]
----
\((1 - \tan \alpha) \cos \alpha = \sin(90 + \alpha) + \cos(90 + \alpha)\)
----
\[\begin{align}
    \text{LHS} &= (1 - \frac{\sin \alpha}{\cos \alpha}) . \cos \alpha \\
    &= (\frac{\cos \alpha - \sin \alpha}{\cos \alpha} ) \cos \alpha \\
    &= \cos \alpha - \sin \alpha \\
    \text{RHS} &= \cos (-\alpha) + \sin (-\alpha) \\
    &= \cos \alpha - \sin \alpha \\
    &= \text{LHS}
  \end{align}\]
----
\(\tan y + \dfrac{1}{\tan y} = \dfrac{1}{\cos^2y \tan y}\)
----
\[\begin{align}
    \text{LHS} &= \frac{\sin y}{\cos y} + \frac{1}{\frac{1}{\sin y}{\cos y}}\\
    &= \frac{\sin y}{\cos y} + \frac{\cos y}{\sin y}\\
    &= \frac{\sin^2 y + \cos^2 y}{\cos y . \sin y}\\
    &= \frac{1}{\cos y . \sin y}\\
    \text{RHS} &= \frac{1}{\cos^2 y \frac{\sin y}{\cos y}}\\
    &= \frac{\cos y}{\cos^2 y . \sin y}\\
    &= \frac{1}{\cos y \sin y}\\
    &= \text{LHS}
  \end{align}\]
----
\[\begin{align}
    & \frac{(-\sin \theta) . (- \tan \theta)}{(-\sin \theta) . \tan \theta} \\
    & = -1
  \end{align}\]
----
\(12 \tan \theta = 5\)
----
\(\sin \theta\)
----
\(-\frac{5}{13}\)
----
\(\frac{12}{13}\)
----
\(\theta\)
----
\(P\)
----
\(OP = 2\)
----
\(a\)
----
\(b\)
----
\(a =  1\)
----
\(b =  -\sqrt{3}\)
----
\(-\frac{\sqrt{3}}{2}\)
----
\(x\)
----
\(\cos^2x - 4 \sin^2x = 0\)
----
\(\sin^2x + 2 \cos x = -2\)
----
\(A\)
----
\(AT\)
----
\(N\)
----
\(n\)
----
\(G\)
----
\(A\)
----
\(\alpha\)
----
\(G\)
----
\(N\)
----
\(\beta\)
----
\(A\hat{G}N\)
----
\(\alpha\)
----
\(\beta\)
----
\[\begin{align}
    A\hat{G}N &= \alpha - \beta
  \end{align}\]
----
\(A\hat{G}N = \alpha - \beta\)
----
\(\hat{A}\)
----
\(\alpha\)
----
\(\beta\)
----
\(H\)
----
\[H = \frac{n \cos \alpha \sin \beta}{\sin(\alpha - \beta)}\]
----
\(H\)
----
\(ABCD\)
----
\(AB \parallel DC\)
----
\(AC\)
----
\(AD\)
----
\(\triangle ABC\)
----
\(AC = 2A\)
----
\(AF = BF\)
----
\(A\hat{F}B = \alpha\)
----
\(FC = 2AF\)
----
\(\cos \alpha = \frac{1}{4}\)
----
\(\text{Area}={s}^{2}\)
----
\(\text{Area}=b\times h\)
----
\(\text{Area}=\frac{1}{2}b\times h\)
----
\(\text{Area}=\frac{1}{2}\left(a+b\right)\times h\)
----
\(\text{Area}=b\times h\)
----
\(\text{Area}=\pi {r}^{2}\)
----
\(\left(\text{Circumference}=2\pi r\right)\)
----
\(ABCD\)
----
\(ABCD\)
----
\(ABCD\)
----
\(ABCD =\)
----
\(\times\)
----
\(ABCD = 2DC + 2BC\)
----
\(BC\)
----
\(AF \perp BC\)
----
\(O\)
----
\(\pi\)
----
\(\begin{aligned}
			  \text{Volume} &= \text{area of base} \times \text{height} \\
			  &= \text{area of rectangle} \times \text{height} \\
			  &= l \times b \times h
			\end{aligned}\)
----
\(\begin{aligned}
			\text{Volume} &= \text{area of base} \times \text{height} \\
			&= \text{area of triangle} \times \text{height} \\
			&= \left(\frac{1}{2}b\times h\right)\times H
			\end{aligned}\)
----
\(\begin{aligned}
			\text{Volume} &= \text{area of base} \times \text{height} \\
			&= \text{area of circle} \times \text{height} \\
			& = \pi r^2 \times h
			\end{aligned}\)
----
\(\times\)
----
\(\times\)
----
\(AE\)
----
\(FG\)
----
\(AE\)
----
\(G\)
----
\(AE\)
----
\(AG = x\)
----
\(\begin{array}{r@{\;}c@{\;}l}
			  \text{Surface area} &= & \text{area of base} \;+ \\
			  && \text{area of triangular sides} \\
			  &=& \left(\frac{1}{2}b\times {h}_{b}\right)+3\left(\frac{1}{2}b\times {h}_{s}\right) \\
			  &=& \frac{1}{2}b\left({h}_{b}+3{h}_{s}\right)
			\end{array}\)
----
\(\begin{array}{r@{\;}c@{\;}l}
			  \text{Surface area} &=& \text{area of base} \;+ \\
			  && \text{area of walls} \\
			  &=& \pi {r}^{2}+\frac{1}{2}\times 2\pi rh \\
			  &=& \pi r\left(r+h\right)
			\end{array}\)
----
\(\begin{array}{r@{\;}c@{\;}l}
			  \text{Volume} &=& \frac{1}{3}\times \text{area of base} \;\times \\
			  &&  \text{height of pyramid} \\
			  &=& \frac{1}{3}\times {b}^{2}\times H
			\end{array}\)
----
\(\begin{array}{r@{\;}c@{\;}l}
			  \text{Volume} &=& \frac{1}{3}\times \text{area of base} \;\times \\
			  && \text{height of pyramid} \\
			  &=& \frac{1}{3}\times \frac{1}{2}bh\times H
			\end{array}\)
----
\(\begin{array}{r@{\;}c@{\;}l}
			  \text{Volume} &=& \frac{1}{3}\times \text{area of base} \;\times \\
			  && \text{height of cone} \\
			  &=&  \frac{1}{3}\times \pi {r}^{2}\times H
			\end{array}\)
----
\(\text{Volume}=\frac{4}{3}\pi {r}^{3}\)
----
\(R\)
----
\(P\)
----
\(R\)
----
\(P\)
----
\(M\)
----
\(l\)
----
\(b\)
----
\(h\)
----
\(\begin{aligned}
		        V &= l \times b \times h \\
			&  =lbh  
		      \end{aligned}\)
----
\(\begin{aligned}
		      A & =2\left[\left(l\times h\right)+\left(l\times b\right)+\left(b\times h\right)\right]\\
		      & =2\left(lh+lb+bh\right)
		      \end{aligned}\)
----
\(\begin{aligned}
		      {V}_{1}&= l\times b\times 5h \\
		      &= 5\left(lbh\right) \\
		      &= 5V
		      \end{aligned}\)
----
\(\begin{aligned}
		        A_1 &= 2\left[\left(l\times 5h\right)+\left(l\times b\right)+\left(b\times 5h\right)\right] \\
			&= 2\left(5lh+lb+5bh\right)
		      \end{aligned}\)
----
\(\begin{aligned}
  V &= 5l \times b \times 5h \\
  &= 5 \dottimes 5 (lbh) \\
  &= 5^2 V
\end{aligned}\)
----
\(\begin{aligned}
  & A_2 \\
  =& 2\left[ (5l\times 5h) + (5l\times b) + (b\times 5h) \right] \\
  =& 2\times 5 (5lh + lb + bh)
\end{aligned}\)
----
\(\begin{aligned}
  V &= 5l \times 5b \times 5h \\
  &= 5^3 (lbh) \\
  &= 5^3 V
\end{aligned}\)
----
\(\begin{aligned}
  & A_3 \\
  =& 2\left[ (5l\times 5h) + (5l\times 5b) + (5b\times 5h) \right] \\
  =& 2\times (5^2lh + 5^2lb + 5^2bh) \\
  =& 5^2 \times 2 (lh + lb + bh) \\
  =& 5^2 A
\end{aligned}\)
----
\(k\)
----
\(\begin{aligned}
  V &= kl \times kb \times kh \\
  &= k^3 (lbh) \\
  &= k^3 V
\end{aligned}\)
----
\(\begin{aligned}
  & A_k \\
  =& 2\left[ (kl\times kh) + (kl\times kb) + (kb\times kh) \right] \\
  =& 2\times (k^2lh + k^2lb + k^2bh) \\
  =& k^2 \times 2 (lh + lb + bh) \\
  =& k^2 A
\end{aligned}\)
----
\(k\)
----
\(k\)
----
\(\sqrt{2}\)
----
\[\begin{align}
		  \text{Area } O &=  \text{length} \times \text{length} \\
		  &= k^2
		  \end{align}\]
----
\(2k^2\)
----
\[\begin{align}
		  \text{Area } M &=  \text{length} \times \text{length} \\
		  &=  2k \times 2k \\
		  &= 4k^2
		  \end{align}\]
----
\[\begin{align}
		  \text{Area } D &=  \text{length} \times \text{length} \\
		  &=  (k+2) \times (k+2) \\
		  &= k^2 + 4k + 2 \\
		  &\ne 2k^2
		  \end{align}\]
----
\(\sqrt{2}\)
----
\[\begin{align}
		  \text{Area } d &=  \text{length} \times \text{length} \\
		  &=  \sqrt{2}k \times \sqrt{2}k \\
		  &= 2k^2
		  \end{align}\]
----
\[\begin{align}
		  \text{Area } s &=  \text{length} \times \text{length} \\
		  &=  2k \times k \\
		  &= 2k^2
		  \end{align}\]
----
\(\sqrt{2}\)
----
\(k\)
----
\(\frac{1}{2}\)
----
\(\ldots\)
----
\(\ldots\)
----
\[\begin{aligned} 
\text{multiplied by } 7^2 &= 49 \\
\therefore &\approx \text{ 50 times bigger}
\end{aligned}\]
----
\(\ldots\)
----
\(\ldots\)
----
\(3^2 = 9\)
----
\(\ldots\)
----
\(3^3 = 27\)
----
\(\ldots\)
----
\(\frac{1}{2}^2 = \frac{1}{4}\)
----
\(\ldots\)
----
\(\frac{1}{2}^3 = \frac{1}{8}\)
----
\(W^3\)
----
\(W\)
----
\(W\)
----
\(\frac{9}{10}\)
----
\({s}^{2}\)
----
\(b\times h\)
----
\(\frac{1}{2}b\times h\)
----
\(\frac{1}{2}\left(a+b\right)\times h\)
----
\(b\times h\)
----
\(\pi {r}^{2}\)
----
\(l\times b\times h\)
----
\(\left(\frac{1}{2}b\times h\right)\times H\)
----
\({s}^{3}\)
----
\(\pi {r}^{2}\times h\)
----
\(b\left(b+2h\right)\)
----
\(\frac{1}{2}b\left({h}_{b}+3{h}_{s}\right)\)
----
\(\pi r\left(r+{h}_{s}\right)\)
----
\(4\pi {r}^{2}\)
----
\(\frac{1}{3}\times {b}^{2}\times H\)
----
\(\frac{1}{3}\times \frac{1}{2}bh\times H\)
----
\(\frac{1}{3}\times \pi {r}^{2}\times H\)
----
\(\frac{4}{3}\pi {r}^{3}\)
----
\(ABCD\)
----
\(\frac{3}{2}x\)
----
\(O\)
----
\(DO = x\)
----
\(ABCD\)
----
\(x\)
----
\(H\)
----
\(h\)
----
\(k\)
----
\[\begin{aligned} 
\text{Volume }&=k \times k\times k \\
&=k^3 \\
\text{New volume }&=k\times k\times 3k \\
&=3k^3
\end{aligned}\]
----
\[\begin{aligned} 
\text{Surface area } &= 6\times k^2 \\
&=6k^2 \\
\text{Surface area }&= 6\times (3k)^2 \\
&=54k^2
\end{aligned}\]
----
\(\times 9\)
----
\[\begin{aligned} 
\text{Volume }&=k^3 \\
\text{New volume }&=(3k)^3 \\
&=27k^3
\end{aligned}\]
----
\(\times 27\)
----
\(AD = DB\)
----
\(AE = EC\)
----
\(DE \parallel BC\)
----
\(DE = \frac{1}{2}BC\)
----
\(r\)
----
\[(AC)^2 = (AB)^2 + (BC)^2\]
----
\(OT \perp ST\)
----
\(A\)
----
\(B\)
----
\(B\)
----
\(A\)
----
\(\perp\)
----
\(O\)
----
\(OP\)
----
\(AB\)
----
\(AP = PB\)
----
\(OA\)
----
\(OB\)
----
\(\triangle OPA\)
----
\(\triangle OPB\)
----
\[\begin{array}{rll}
      OA^2 &= OP^2 + AP^2 & \text{(Pythagoras)} \\
      OB^2 &= OP^2 + BP^2 & \text{(Pythagoras)}
      \end{array}\]
----
\[\begin{array}{rll}
      OA  &= OB & \text{(equal radii)} \\
      \therefore AP^2 &= BP^2 & \\
      \therefore AP &= BP &
      \end{array}\]
----
\(OP\)
----
\(AB\)
----
\(\triangle OPA\)
----
\(\triangle OPB\)
----
\[\begin{array}{rll}
      O\hat{P}A &= O\hat{P}B & (\text{given } OP \perp AB) \\
      OA &= OB & \text{(equal radii)} \\
      OP &= OP & \text{(common side)}  \\
      \therefore \triangle OPA & \equiv  \triangle OPB & \text{(RHS)}  \\
      \therefore AP &= PB &
      \end{array}\]
----
\(OP\)
----
\(AB\)
----
\(\perp\)
----
\(O\)
----
\(OP\)
----
\(P\)
----
\(AB\)
----
\(OP \perp AB\)
----
\(OA\)
----
\(OB\)
----
\(\triangle OPA\)
----
\(\triangle OPB\)
----
\(OP \perp AB\)
----
\(\perp\)
----
\(P\)
----
\(AB\)
----
\(QP\)
----
\(RP\)
----
\(O\)
----
\(PR\)
----
\(QA\)
----
\(QB\)
----
\(RA\)
----
\(RB\)
----
\(\triangle QPA\)
----
\(\triangle QPB\)
----
\(\triangle RPA\)
----
\(\triangle RPB\)
----
\(RA = RB\)
----
\(A\)
----
\(B\)
----
\(PR\)
----
\(O\)
----
\(PR\)
----
\(OQ \perp PR\)
----
\(PR = 8\)
----
\(x\)
----
\(PQ = QR = 4 \qquad (\perp \text{ from centre bisects chord})\)
----
\(x\)
----
\(\triangle OQP\)
----
\[\begin{array}{rll}
        PQ &= 4   & (\perp \text{ from centre bisects chord}) \\
        OP^2 &= OQ^2 + QP^2 & (\text{Pythagoras}) \\
        5^2 &= x^2 + 4^2   & \\
        \therefore x^2 &= 25 - 16  & \\
        x^2 &= 9  & \\
        x &= 3 &
        \end{array}\]
----
\(x = 3\)
----
\(O\)
----
\(OQ \perp PR\)
----
\(OQ = 4\)
----
\(PR=10\)
----
\(x\)
----
\[\begin{array}{rll}
        PR &= 10   & (\perp \text{ given }) \\
        \therefore         PQ &= 5   & (\perp \text{ from centre bisects chord}) \\
        OP^2 &= OQ^2 + QP^2 & (\text{Pythagoras}) \\
        x^2 &= 4^2 + 5^2   & \\
        \therefore x^2 &= 25 + 16  & \\
        x^2 &= 41  & \\
        x &= \sqrt{41} &
        \end{array}\]
----
\(x=\sqrt{41}\)
----
\(O\)
----
\(= 10\)
----
\(OQ \perp PR\)
----
\(PR=8\)
----
\(x\)
----
\[\begin{array}{rll}
        PR &= 8   & (\perp \text{ given }) \\
        \therefore         PQ &= 4   & (\perp \text{ from centre bisects chord}) \\
        OP^2 &= OQ^2 + QP^2 & (\text{Pythagoras}) \\
        10^2 &= x^2 + 4^2   & \\
        \therefore x^2 &= 100 - 16  & \\
        x^2 &= 84 & \\
        x &= \sqrt{84} &
        \end{array}\]
----
\(x=\sqrt{84}\)
----
\(O\)
----
\(OQ \perp PR\)
----
\(PR = 12\)
----
\(SQ = 2\)
----
\(x\)
----
\[\begin{array}{rll}
        SO &= x -2   & \\
      OP^2 &= OS^2 + SP^2 & (\text{Pythagoras}) \\
        x^2 &= (x-2)^2 + 6^2   & \\
        x^2 &= x^2 - 4x + 4 + 6^2   & \\
        4x &= 40   & \\
        \therefore x &= 10 &
        \end{array}\]
----
\(x=10\)
----
\(O\)
----
\(OT \perp SQ\)
----
\(OT \perp PR\)
----
\(OP = 10\)
----
\(ST = 5\)
----
\(PU = 8\)
----
\(TU\)
----
\(O\)
----
\(OT \perp QP\)
----
\(OS \perp PR\)
----
\(OT = 5\)
----
\(PQ = 24\)
----
\(PR = 25\)
----
\(OS = x\)
----
\(x\)
----
\(y\)
----
\(x\)
----
\(y\)
----
\(\angle \text{ at centre } = 2 \angle \text{ at circum.}\)
----
\(O\)
----
\(AB\)
----
\(A\hat{O}B\)
----
\(A\hat{P}B\)
----
\(A\hat{O}B= 2A\hat{P}B\)
----
\(PO\)
----
\(Q\)
----
\(A\hat{O}Q = \hat{O}_1\)
----
\(B\hat{O}Q = \hat{O}_2\)
----
\[\begin{array}{rll}
      \hat{O}_1&= A\hat{P}O  + P\hat{A}O    & (\text{ext. } \angle \triangle = \text{sum int. opp. } \angle \text{s} ) \\
      \text{and } A\hat{P}O &=  P\hat{A}O     & (\text{equal radii, isosceles } \triangle APO)\\
      \therefore \hat{O}_1&= A\hat{P}O  +  A\hat{P}O   &  \\
        \hat{O}_1&= 2A\hat{P}O     &  
\end{array}\]
----
\(\hat{O}_2 = 2B\hat{P}O\)
----
\[\begin{array}{rll}
     A\hat{O}B  &= \hat{O}_1 + \hat{O}_2  &  \\
                         &= 2A\hat{P}O  + 2B\hat{P}O    &  \\
                         &= 2(A\hat{P}O  + B\hat{P}O)    &  \\
  \therefore A\hat{O}B &= 2(A\hat{P}B)    &  
\end{array}\]
----
\[\begin{array}{rll}
     A\hat{O}B  &= \hat{O}_2 - \hat{O}_1  &  \\
                         &= 2B\hat{P}O  - 2A\hat{P}O    &  \\
                         &= 2(B\hat{P}O  - A\hat{P}O)    &  \\
  \therefore A\hat{O}B &= 2(A\hat{P}B)    &  
\end{array}\]
----
\(HK\)
----
\(O\)
----
\(a\)
----
\(\triangle HJK\)
----
\(O\)
----
\(a\)
----
\(b\)
----
\(c\)
----
\(d\)
----
\(e\)
----
\(A\)
----
\(B\)
----
\(AP\)
----
\(BP\)
----
\(A\hat{P}B\)
----
\(AQ\)
----
\(BQ\)
----
\(A\hat{Q}B\)
----
\(\angle\)
----
\(O\)
----
\(P\)
----
\(Q\)
----
\(AB\)
----
\(A\hat{P}B\)
----
\(A\hat{Q}B\)
----
\(A\hat{P}B = A\hat{Q}B\)
----
\[\begin{array}{rll}
      A\hat{O}B &=  2 A\hat{P}B  & (\angle \text{ at centre } = 2 \angle \text{ at circum.}) \\
      A\hat{O}B &=  2 A\hat{Q}B  & (\angle \text{ at centre } = 2 \angle \text{ at circum.}) \\
      \therefore 2 A\hat{P}B &= 2 A\hat{Q}B    &  \\
         A\hat{P}B &=  A\hat{Q}B    &  
\end{array}\]
----
\(AB\)
----
\(P\)
----
\(Q\)
----
\(AB\)
----
\(A\)
----
\(B\)
----
\(P\)
----
\(Q\)
----
\(P\)
----
\(AP\)
----
\(R\)
----
\(A\)
----
\(B\)
----
\(Q\)
----
\(P\)
----
\(A\)
----
\(B\)
----
\(Q\)
----
\(P\)
----
\(A\)
----
\(B\)
----
\(Q\)
----
\(P\)
----
\(FH \parallel EI\)
----
\(b\)
----
\(b\)
----
\(T\hat{V}S = S\hat{V}R\)
----
\(e\)
----
\(TV\)
----
\(O\)
----
\(WT = TY\)
----
\(f\)
----
\(ABCD\)
----
\(\ldots \ldots\)
----
\(\hat{A} =\)
----
\(\hat{B} =\)
----
\(\hat{C} =\)
----
\(\hat{D} =\)
----
\(\hat{A} + \hat{C} =\)
----
\(\hat{B} + \hat{D} =\)
----
\(\angle\)
----
\(O\)
----
\(A, B, P\)
----
\(Q\)
----
\(ABPQ\)
----
\(AO\)
----
\(OP\)
----
\(\hat{O}_1\)
----
\(\hat{O}_2\)
----
\(O\)
----
\(PQRS\)
----
\(SQ\)
----
\(a\)
----
\(b\)
----
\(c\)
----
\(b\)
----
\(\triangle PSQ\)
----
\(PQRS\)
----
\(\hat{P} = \hat{Q}\)
----
\(\hat{S} = \hat{R}\)
----
\(PQRS\)
----
\(T\hat{Q}R = \hat{S}\)
----
\(PQRS\)
----
\(ABDE\)
----
\(ABDE\)
----
\(ABCD\)
----
\(O\)
----
\(PA\)
----
\(PB\)
----
\(A\)
----
\(B\)
----
\(AP = BP\)
----
\(\triangle AOP\)
----
\(\triangle BOP\)
----
\(a\)
----
\(b\)
----
\(c\)
----
\(a\)
----
\(b\)
----
\(c\)
----
\[\begin{array}{rll}
          AB = AF&= a  & (\text{tangents from } A) \\
          EF = ED&= c  & (\text{tangents from } E) \\
          CB = CD&= b  & (\text{tangents from } C) \\
          \therefore AE = a + c &= 5 & \\
          \text{and } AC = a + b &= 8 & \\
          \text{and } CE = b + c &= 9 &
  \end{array}\]
----
\[\begin{array}{rll}
        a + c &= 5 & \ldots (1)\\
        a + b &= 8 & \ldots (2)\\
        b + c &= 9 & \ldots (3)
  \end{array}\]
----
\((1)\)
----
\((2)\)
----
\((3)\)
----
\[\begin{array}{rll}
        (2) - (1) \quad b-c &= 8-5 & \\
        &= 3 & \\
        \therefore b &= c + 3 & \\
        \text{Substitute into } (3) \quad c + 3 + c &= 9 & \\
        2c &= 6 & \\
        c &= 3 & \\
        \therefore a &= 2 & \\
        \text{ and } b &= 6 &
  \end{array}\]
----
\(A\hat{B}C =\)
----
\(\hat{D} =\)
----
\(\hat{E} =\)
----
\(\ldots \ldots\)
----
\(O\)
----
\(SR\)
----
\(B\)
----
\(AB\)
----
\(\hat{P}_1\)
----
\(\hat{Q}_1\)
----
\(A\hat{B}R = A\hat{P}B\)
----
\(A\hat{B}S = A\hat{Q}B\)
----
\(BT\)
----
\(T\)
----
\(A\)
----
\(A\hat{T}B = T_1\)
----
\(h\)
----
\(s\)
----
\(h\)
----
\(s\)
----
\(O\)
----
\(SPT\)
----
\(OP \perp ST\)
----
\(a\)
----
\(b\)
----
\(c\)
----
\(AB = AC\)
----
\(AP \parallel BC\)
----
\(\hat{A}_2 = \hat{B}_2\)
----
\(PAL\)
----
\(ABC\)
----
\[\begin{array}{rll}
        \hat{A}_1 &=  A\hat{C}B  & (\text{alt. angles}, AP \parallel BC) \\
                A\hat{C}B &=  A\hat{B}C & (\angle \text{s opp. equal sides }, AB = AC) \\
        \therefore \hat{A}_1  &=   A\hat{B}C  &   \\
        \text{Therefore } PAL &  \text{is a tangent to circle }  ABC   & (\angle\text{ between line chord} = \angle\text{ in alt. seg.})
     \end{array}\]
----
\(AB\)
----
\(ADP\)
----
\[\begin{array}{rll}
        \hat{A}_2 &=  \hat{B}_2  & (\text{ given }) \\
      \text{And } A\hat{P}B  &=    \hat{B}_2  &   (\text{alt. angles}, AP \parallel BC) \\
        \text{Therefore } A\hat{P}B &=  \hat{A}_2  ABC   &  \\
        \text{Therefore } AB &  \text{is a tangent to circle }  ADP   & (\angle\text{ between line chord} = \angle\text{ in alt. seg.})
     \end{array}\]
----
\(\angle\)
----
\(= \angle\)
----
\(BD\)
----
\(O\)
----
\(BO \perp AD\)
----
\(CFOE\)
----
\(FB=BC\)
----
\(\angle A\hat{O}C = 2 B\hat{F}C\)
----
\(DC\)
----
\(C,F,O\)
----
\(E\)
----
\(CFOE\)
----
\(BFC\)
----
\(FB = BC\)
----
\(\triangle BFC\)
----
\(B\hat{F}C = B\hat{C}F\)
----
\[\begin{array}{rll}
               B\hat{C}F &= C\hat{E}O & (\text{tangent-chord}) \\
               C\hat{E}O &= B\hat{F}C & (\text{ext. } \angle \text{ cyclic quad. } CFOE) \\
               \therefore B\hat{F}C &= B\hat{C}F \\
               \therefore FB &= BC & (\triangle BFC \text{ isosceles})
       \end{array}\]
----
\(A\hat{O}C = 2 B\hat{F}C\)
----
\[\begin{array}{rll}
              A\hat{O}C &= 2 A\hat{E}C & (\angle \text{ at centre } = 2 \angle \text{ at circum.}) \\
              \text{and } A\hat{E}C &= B\hat{F}C    & (\text{ext. } \angle \text{ cyclic quad. } CFOE)  \\
              \therefore A\hat{O}C &= 2 B\hat{F}C
      \end{array}\]
----
\(DC\)
----
\(C\)
----
\(F\)
----
\(O\)
----
\(E\)
----
\(DC\)
----
\(C\)
----
\(F\)
----
\(O\)
----
\(E\)
----
\[\begin{array}{rll}
              \therefore D\hat{C}E = C\hat{O}E \quad (\text{tangent-chord})
        \end{array}\]
----
\(O\)
----
\(BD\)
----
\[\begin{array}{rll}
              D\hat{C}E &= C\hat{A}E    & (\text{tangent-chord}) \\
              \text{but } C\hat{A}E &= \frac{1}{2} C\hat{O}E & (\angle \text{ at centre} = 2 \angle \text{ at circum.}) \\
              \therefore D\hat{C}E &\ne C\hat{O}E &
        \end{array}\]
----
\(DC\)
----
\(C\)
----
\(F\)
----
\(O\)
----
\(E\)
----
\(FD\)
----
\(CB\)
----
\(FADE\)
----
\(F\hat{E}A = \hat{B}\)
----
\(FADE\)
----
\[\begin{array}{rll}
              F\hat{D}C &= D\hat{C}B &  (\text{alt. } \angle\text{s } FD \parallel CB) \\
              \text{and } D\hat{C}B  &= C\hat{A}E    &  (\text{tangent-chord})   \\
              \therefore F\hat{D}C &= C\hat{A}E \\
              \therefore FADE & \text{is a cyclic quad.}   & (\angle \text{s in same seg.})
      \end{array}\]
----
\(F\hat{E}A = \hat{B}\)
----
\[\begin{array}{rll}
              F\hat{D}A &= \hat{B} & (\text{corresp. } \angle\text{s } FD \parallel CB) \\
              \text{and } F\hat{E}A  &= F\hat{D}A    &  (\angle \text{s same seg.\@ cyclic quad.\@ } FADE)   \\
              \therefore F\hat{E}A &= \hat{B}
      \end{array}\]
----
\(r\)
----
\(O\)
----
\(OM \perp AB\)
----
\(AM = MB\)
----
\(O\)
----
\(AM = MB\)
----
\(AM = MB\)
----
\(OM \perp AB\)
----
\(\Rightarrow MO\)
----
\(O\)
----
\(ABCD\)
----
\(O\)
----
\(\angle\)
----
\(\angle\)
----
\(E\hat{B}C = \hat{D}\)
----
\(\angle\)
----
\(\hat{A}_1 = \hat{A}_2 = \hat{C}\)
----
\(\angle\)
----
\(\angle\)
----
\(ABCD\)
----
\(\hat{A}_1 = \hat{C}\)
----
\(\hat{D}_1 = \hat{B}\)
----
\(ABCD\)
----
\(\hat{A} = \hat{B}\)
----
\(\hat{C} = \hat{D}\)
----
\(ABCD\)
----
\(AT\)
----
\(BT\)
----
\(O\)
----
\(OA \perp AT\)
----
\(\perp\)
----
\(OB \perp BT\)
----
\(\perp\)
----
\(TA = TB\)
----
\(DC\)
----
\(D\hat{T}A = T\hat{B}A\)
----
\(C\hat{T}B = T\hat{A}B\)
----
\(D\hat{T}A = T\hat{B}A\)
----
\(C\hat{T}B = T\hat{A}B\)
----
\(DC\)
----
\(T\)
----
\(AOC\)
----
\(O\)
----
\(F\)
----
\(EC\)
----
\(B\hat{O}C = C\hat{O}D\)
----
\(\hat{B} = x\)
----
\(x\)
----
\(\hat{A}\)
----
\[\begin{array}{rll}
    \hat{A} &= x & (\text{radius } OA = OB)
  \end{array}\]
----
\(\hat{A} = x\)
----
\(C\hat{O}D\)
----
\(C\hat{O}D = 2x\)
----
\(\hat{D}\)
----
\(D\)
----
\(E\)
----
\(F\)
----
\(G\)
----
\(M\)
----
\(\hat{M}_1\)
----
\(\hat{D}_1\)
----
\(\hat{F}_2\)
----
\(\hat{G}\)
----
\(\hat{E}_1\)
----
\(O\)
----
\(M\)
----
\(O\)
----
\(DA\)
----
\(C\)
----
\(\hat{D}_1 = x\)
----
\(x\)
----
\(\hat{D}_2\)
----
\[\begin{array}{rll}
    \hat{D_2} &= \hat{D_1} &(\angle\text{s on same chord} OA = OB) \\
    &= x & 
  \end{array}\]
----
\(\hat{D}_{2} = x\)
----
\(O\hat{A}B\)
----
\[\begin{array}{rll}
    O\hat{A}B &= x &(\angle\text{'s on same chord}) 
  \end{array}\]
----
\(O\hat{A}B = x\)
----
\(O\hat{B}A\)
----
\[\begin{array}{rll}
    O\hat{B}A &= x &(\text{equal radii} OA=OB) 
  \end{array}\]
----
\(O\hat{B}A = x\)
----
\(A\hat{O}B\)
----
\(\hat{C}\)
----
\(O\)
----
\(OM\)
----
\(AM\)
----
\(AB\)
----
\(AO \parallel CB\)
----
\(O\)
----
\(O\hat{A}C = x\)
----
\(x\)
----
\(PQ\)
----
\(O\)
----
\(SQ\)
----
\(P\hat{Q}R\)
----
\(P\hat{Q}S = x\)
----
\(x\)
----
\[\begin{array}{rll}
    R\hat{Q}S &= x &  \\
    Q\hat{S}O &= x & 
  \end{array}\]
----
\(R\hat{Q}S\)
----
\(Q\hat{S}O\)
----
\(P\hat{O}S\)
----
\(x\)
----
\(P\hat{O}S = 2x\)
----
\(OS\)
----
\(PR\)
----
\(B\hat{O}D\)
----
\(O\)
----
\(AB = AD\)
----
\(O\hat{D}C\)
----
\(C\hat{O}D\)
----
\(C\hat{B}D\)
----
\(B\hat{A}D\)
----
\(A\hat{D}B\)
----
\(QP\)
----
\(O\)
----
\(T\)
----
\(PR = PT\)
----
\(y\)
----
\(x\)
----
\(x = 4y\)
----
\(O\)
----
\(AB\)
----
\(CD \perp AB\)
----
\(P\)
----
\(DE\)
----
\(AB\)
----
\(F\)
----
\(C\hat{B}P = D\hat{P}B\)
----
\(C\hat{E}D = 2 C\hat{B}A\)
----
\[\begin{array}{rll}
    C\hat{E}D &= C\hat{B}D& (\angle\text{'s on chord } CD)\\
    \text{But } C\hat{B}A &= D\hat{B}A & (\triangle CBP \equiv \triangle DBP)\\
    \therefore C\hat{E}D &= 2 C\hat{B}A
  \end{array}\]
----
\(A\hat{B}D = \frac{1}{2} C\hat{O}A\)
----
\[\begin{array}{rll}
    D\hat{B}A &= C\hat{B}A & (\triangle CBP \equiv \triangle DBP)\\
    C\hat{B}A &=  \frac{1}{2}  C\hat{O}A & (\angle\text{'s at centre and circumference on arc } AC)\\
    \therefore A\hat{B}D &= \frac{1}{2} C\hat{O}A &
  \end{array}\]
----
\(O\)
----
\(OR \perp QP\)
----
\(OQ\)
----
\[\begin{array}{rll}
    \text{In } &\triangle QOS,  & \\
    QO^2 &=  OS^2 + QS^2 & ( \text{Pythagoras }) \\
    x^2 &=  (x-9)^2 + 15^2 &  \\
    x^2 &=  x^2 - 18x + 81  + 225 &  \\
    \therefore 18x &=   306  &  \\
    \therefore x &=   17  &  
  \end{array}\]
----
\(P\)
----
\(Q\)
----
\(R\)
----
\(S\)
----
\(M\)
----
\(PS\)
----
\(QR\)
----
\(T\)
----
\(PQ = PR\)
----
\(Q\hat{R}P,Q\hat{S}P,R\hat{S}T\)
----
\(S\hat{R}T\)
----
\(S\hat{T}R\)
----
\(P\hat{Q}S\)
----
\(PQ\)
----
\(QST\)
----
\(Q\)
----
\(PQ\)
----
\(QST\)
----
\(P\hat{M}Q\)
----
\(POQ\)
----
\(O\)
----
\(QP\)
----
\(A\)
----
\(AC\)
----
\(BA \perp AQ\)
----
\(BCQ\)
----
\(P\hat{C}Q = B\hat{A}P\)
----
\(BAPC\)
----
\[\begin{array}{rll}
      P\hat{C}Q &= B\hat{A}Q & ( \text{ proven } ) \\
      \therefore &BAPC \text{ is a cyclic quad. } &  (  \text{ ext. angle = int. opp. angle } )  
    \end{array}\]
----
\(AB = AC\)
----
\[\begin{array}{rll}
       C\hat{P}Q &= A\hat{B}C & ( \text{ ext. angle of cyclic quad. } ) \\
       B\hat{C}P &= C\hat{P}Q + C\hat{Q}P   &  (  \text{ ext. angle of } \triangle )  \\
       A\hat{C}P &=  C\hat{Q}P   &  ( \text{ tangent-chord } )  \\
                \therefore  B\hat{C}A &=  C\hat{P}Q  &   \\
                \therefore    &=  A\hat{B}C  &   \\
                \therefore   AB  &= AC & \text{ angles opp. equal sides}
    \end{array}\]
----
\(TA\)
----
\(TB\)
----
\(O\)
----
\(C\)
----
\(A\hat{T}B = x\)
----
\(x\)
----
\(A\hat{B}T\)
----
\(O\hat{B}A\)
----
\(O\hat{B}A = \frac{x}{2}\)
----
\(\hat{C}\)
----
\(AOB\)
----
\(AECB\)
----
\(O\)
----
\(OE \parallel BC\)
----
\(AC\)
----
\(D\)
----
\(AD = DC\)
----
\(A\hat{B}C\)
----
\(EB\)
----
\[\begin{array}{rll}
       O\hat{E}B &= E\hat{B}C    & ( \text{ alt. } \angle, OE \parallel BC ) \\
       OE &= OB  &  (\angle \text{ equal radii} )  \\
       \therefore O\hat{E}B &=  O\hat{B}E  &  (\angle \text{ opp. equal sides }  )  \\
                   \therefore  O\hat{B}E   &=  E\hat{B}C  &  \\
                   \therefore  A\hat{B}C   &\text{ is bisected}  &
     \end{array}\]
----
\(O\hat{E}B = x\)
----
\(B\hat{A}C\)
----
\(x\)
----
\(PQ\)
----
\(RS\)
----
\(PQ \parallel RS\)
----
\(Q\)
----
\(RS\)
----
\(T\)
----
\(S\)
----
\(QT\)
----
\(V\)
----
\(QS\)
----
\(PR\)
----
\(T\hat{Q}S = x\)
----
\(Q\hat{R}P = y\)
----
\(T\hat{V}S = 2 Q\hat{R}S\)
----
\[\begin{array}{rll}
       \text{In }   &\triangle VQS,    &  \\
                VQ &= VS    &  ( \text{ tangents from same pt. } ) \\
        \therefore V\hat{S}Q  &= V\hat{Q}S = x  &  (\angle \text{s opp. equal sides  })  \\
                \therefore T\hat{V}S &= x + x   & (\text{ext. angle of } \triangle ) \\
         &= 2x  &  \\
                    \text{And } Q\hat{R}S  &= Q\hat{P}S = x &  (\angle \text{ same seg. })  \\
                \therefore T\hat{V}S &= 2Q\hat{R}S   &
    \end{array}\]
----
\(QVSW\)
----
\[\begin{array}{rll}
                T\hat{V}S &= 2Q\hat{R}S    &  ( \text{ proven }  )\\
                        V\hat{Q}S  &= Q\hat{P}S    &  ( \text{ tangent-chord }  )\\
                         Q\hat{P}S    &= P\hat{S}R = x    &  ( \text{ alt. } \angle, PQ \parallel RT  )\\
                    \text{And } Q\hat{W}S  &= 2Q\hat{R}S  &  (\angle \text{ ext. angle of } \triangle WSR )  \\
                    \therefore T\hat{V}S &= Q\hat{W}S    &= 2x  \\
                 \therefore QVSW &= \text{ is cyclic quad. }   &   \text{ext. angle = int. opp. angle }
    \end{array}\]
----
\(Q\hat{P}S + \hat{T} = P\hat{R}T\)
----
\[\begin{array}{rll}
                        \hat{Q} &= Q\hat{R}P = y   &  ( \text{ tangent-chord }  )\\
                \hat{Q} &= \hat{T} = y    &  ( \text{ alt. } \angle, PQ \parallel RT  )\\
                         P\hat{R}T    &= x + y    &   \\
                    \text{And } Q\hat{P}S  &= x  &  (\angle \text{ proven }  )  \\
                    \therefore Q\hat{P}S + \hat{T} &= x + y   &= P\hat{R}T
    \end{array}\]
----
\(W\)
----
\[\begin{array}{rll}
                 Q\hat{R}S  &= x    &  ( \text{ proven }  )\\
                        Q\hat{W}S  &= 2x    &  ( \text{ proven }  )\\
                 \therefore W &= \text{ is circle centre. }   &   (\angle \text{ centre = 2 angle at circum. }
    \end{array}\]
----
\(F\)
----
\(D\)
----
\(BFT\)
----
\(F\)
----
\(AFE\)
----
\(DF = EF\)
----
\(CDE\)
----
\(AC\)
----
\(BF\)
----
\(K\)
----
\(BT \parallel CE\)
----
\[\begin{array}{rll}
                 \hat{E} &= \hat{D}_2    &  (\angle \text{s opp. equal sides }  )\\
                        \hat{F}_4  &= \hat{F}_1    &  ( \text{ vert. opp angles }  )\\
                        \text{ and } \hat{F}_4  &= \hat{D}_2    &  ( \text{ tangent-chord}  )\\
                        \therefore \hat{F}_1  &= \hat{D}_2    &   \\
                 \therefore  BT &\parallel CE  &   (\angle \text{ corresp. angles }
    \end{array}\]
----
\(BCEF\)
----
\[\begin{array}{rll}
                 \therefore  BT & \parallel CE  &   ( \text{ proven}) \\
                        \therefore \hat{F}_1  &= \hat{C}_1    &   ( \text{ angles same seg.}) \\
                 \therefore  AE &\parallel BC  &   (\angle \text{ alt. angles } \\
                 \therefore  BCEF &\text{ is parallelogram  } &   (\angle \text{ both opp. sides } \parallel
    \end{array}\]
----
\(AC = BF\)
----
\[\begin{array}{rll}
                     \hat{F}_1  &= \hat{C}_1    &   ( \text{ proven }) \\
                     \hat{F}_1  &= \hat{E}     &   ( \text{ proven }) \\
                     \hat{A}   &= \hat{C}_1    &   ( \text{ alt. angles }, AE \parallel BC ) \\
                        \therefore \hat{A}  &= \hat{E}    &    \\
                 \therefore  AC &= CE  &   (\angle \text{ angles opp. equal sides } \\
                   BF &= CE  &   (\angle \text{ opp. sides parm. =} \\
                   \therefore AC &= BF  &
    \end{array}\]
----
\[A = P(1 + in)\]
----
\[A = P(1 + i)^n\]
----
\[\begin{align}
	      A &= \text{accumulated amount} \\
	      P &= \text{principal amount}  \\
	      i &= \text{interest rate written as a decimal} \\
	      n &= \text{time period in years}
	    \end{align}\]
----
\(i\)
----
\(i\)
----
\[\text{Total depreciation} = P \times i \times n\]
----
\[\begin{align}
	P &= \text{principal amount}  \\
	i &= \text{interest rate written as a decimal} \\
	n &= \text{time period in years}
	\end{align}\]
----
\[A = P(1 - in)\]
----
\(A = P(1 + in)\)
----
\[A = P(1 - in)\]
----
\(A\)
----
\[A = P(1 - i)^n\]
----
\[\begin{align}
	A &= \text{book value or depreciated value}  \\
	P &= \text{principal amount}  \\
	i &= \text{interest rate written as a decimal} \\
	n &= \text{time period in years}
	\end{align}\]
----
\(A = P(1 + i)^n\)
----
\[A = P(1 - i)^n\]
----
\(A\)
----
\(A\)
----
\(A\)
----
\[A = P(1 - i)^n\]
----
\(P\)
----
\(i\)
----
\(i\)
----
\(x\)
----
\[\begin{align}
						A &= \frac{x}{2} \\
						P &= x \\
						n &= 4
						\end{align}\]
----
\[A = P(1 - in)\]
----
\(i\)
----
\(i\)
----
\[A = P(1 - i)^n\]
----
\(i\)
----
\(i\)
----
\(\frac{1}{6}\)
----
\(p\)
----
\(p\)
----
\(p\)
----
\(p\)
----
\(T_0\)
----
\(T_1\)
----
\(T_4\)
----
\(A\)
----
\(T_1\)
----
\(A\)
----
\(T_4\)
----
\(T_4\)
----
\(T_0\)
----
\(T_0\)
----
\(T_4\)
----
\(T_4\)
----
\(T_6\)
----
\(T_1\)
----
\(T_1\)
----
\(T_4\)
----
\(T_4\)
----
\(T_6\)
----
\(T_5\)
----
\(T_{1\frac{1}{2}}\)
----
\(T_0\)
----
\(T_3\)
----
\[\begin{align}
							A &= P(1 + i)^n   \\
							&=  \ldots \ldots
							 \end{align}\]
----
\(i^{(m)}\)
----
\(m\)
----
\(i\)
----
\[\begin{align}
		  P(1 + i) &=  P \left( 1 + \frac{i^{(m)}}{m} \right)^m \\
		  \therefore 1 + i &=  \left( 1 + \frac{i^{(m)}}{m} \right)^m
		   \end{align}\]
----
\[1 + i =  \left( 1 + \frac{i^{(m)}}{m} \right)^m\]
----
\(i\)
----
\[1 + i =  \left( 1 + \frac{i^{(m)}}{m} \right)^m\]
----
\(i^{(m)}\)
----
\(A = P(1 + in)\)
----
\(A = P(1 + i)^n\)
----
\(A = P(1 - in)\)
----
\(A = P(1 - i)^n\)
----
\(1 + i =  \left( 1 + \frac{i^{(m)}}{m} \right)^m\)
----
\(\left\{1;2;3;4;5;6\right\}\)
----
\(S\)
----
\(A\)
----
\(B\)
----
\(A \text{ or } B\)
----
\(A \cup B\)
----
\(A\)
----
\(B\)
----
\(A \text{ and } B\)
----
\(A \cap B\)
----
\((A \text{ and } B) = \emptyset\)
----
\(A\)
----
\(\text{not } A\)
----
\(A'\)
----
\(A\)
----
\(B\)
----
\[P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\]
----
\[P(A \text{ or } B) = P(A) + P(B)\]
----
\(P(A \text{ and } B) = 0\)
----
\[P(\text{not } A) = 1 - P(A)\]
----
\(A\)
----
\((\text{not } A)\)
----
\(P(A \text{ or } (\text{not } A)) = 1\)
----
\[S = \{\text{A}; 2; 3; 4; 5; 6; 7; 8; 9; 10; \text{J}; \text{Q}; \text{K}\}\]
----
\(\{\text{A}\}\)
----
\(\{2; 3; 5; 7\}\)
----
\(\{\text{A}; \text{J}; \text{Q}; \text{K}\}\)
----
\(\frac{1}{13}\)
----
\(\frac{4}{13}\)
----
\(\frac{4}{13}\)
----
\(E\)
----
\(F\)
----
\(E\)
----
\(F\)
----
\(E\)
----
\(F\)
----
\(E\)
----
\(F\)
----
\(\{1;2;3;4;5;6\}\)
----
\[S = \left\{\begin{array}{cccccc}
  (1;1) & (1;2) & (1;3) & (1;4) & (1;5) & (1;6) \\
  (2;1) & (2;2) & (2;3) & (2;4) & (2;5) & (2;6) \\
  (3;1) & (3;2) & (3;3) & (3;4) & (3;5) & (3;6) \\
  (4;1) & (4;2) & (4;3) & (4;4) & (4;5) & (4;6) \\
  (5;1) & (5;2) & (5;3) & (5;4) & (5;5) & (5;6) \\
  (6;1) & (6;2) & (6;3) & (6;4) & (6;5) & (6;6)
\end{array}\right\}\]
----
\(E\)
----
\[E = \{(4;6); (5;5); (6;4)\}\]
----
\(F\)
----
\[F = \{(1;3);(3;1);(2;3);(3;2);(3;3);(4;3);(3;4);(5;3);(3;5);(6;3);(3;6)\}\]
----
\(6 \times 6 = 36\)
----
\(S\)
----
\(E\)
----
\(F\)
----
\[P(E) = \frac{3}{36} = \frac{1}{12}\]
----
\[P(F) = \frac{11}{36}\]
----
\(E\)
----
\(E\)
----
\(F\)
----
\((E \text{ and } F) = \emptyset\)
----
\(r\)
----
\(b\)
----
\(y\)
----
\(y\)
----
\(r+b+y\)
----
\(\frac{7}{12}\)
----
\(H\)
----
\(T\)
----
\[S = \left\{\begin{array}{cccc}
(H;H;H;H) & (H;H;H;T) & (H;H;T;H) & (H;H;T;T) \\
(H;T;H;H) & (H;T;H;T) & (H;T;T;H) & (H;T;T;T) \\
(T;H;H;H) & (T;H;H;T) & (T;H;T;H) & (T;H;T;T) \\
(T;T;H;H) & (T;T;H;T) & (T;T;T;H) & (T;T;T;T)
\end{array}\right\}\]
----
\[E = \left\{(H;H;T;T); (H;T;H;T); (H;T;T;H); (T;H;H;T); (T;H;T;H); (T;T;H;H)\right\}\]
----
\(E\)
----
\(S\)
----
\(E\)
----
\(F\)
----
\[\begin{align}
		    n(E) &= 15 \\
		    n(F) &= 20 \\
		    n(E \text{ and } F) &= 12
		    \end{align}\]
----
\(n(E \text{ and } (\text{not } F))\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\[\{(2;6); (3;5); (4;4); (5;3); (6;2)\}\]
----
\(A\)
----
\(B\)
----
\(1\)
----
\((A \text{ and } B) = \emptyset\)
----
\(\{14; 28; 42; 56; 70\}\)
----
\(42 - 3 = 39\)
----
\(39 - 23 = 16\)
----
\(S\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(B\)
----
\(A = \{3;6;9;12;15;18\}\)
----
\(B = \{5;10;15;20\}\)
----
\(P(A) = \frac{6}{20} = \frac{3}{10}\)
----
\(P(B) = \frac{4}{20} = \frac{1}{5}\)
----
\(\frac{9}{20}\)
----
\(\frac{1}{20}\)
----
\(M\)
----
\(H\)
----
\(50-12=38\)
----
\((M\text{ or }H)\)
----
\(n(M) = 35\)
----
\(n(H) = 30\)
----
\(n(M\text{ or }H) = 38\)
----
\[\begin{align}
  n(M\text{ or }H) &= n(M) + n(H) - n(M\text{ and }H) \\
  \therefore n(M\text{ and }H) &= 35+30-38 \\
  &= 27
\end{align}\]
----
\(P(M) = \frac{35}{50} = \frac{7}{10}\)
----
\(P(H) = \frac{30}{50} = \frac{3}{5}\)
----
\(P(M\text{ or }H) = \frac{38}{50} = \frac{19}{25}\)
----
\(P(M\text{ and }H) = \frac{27}{50}\)
----
\(P(M\text{ and }H) = \frac{27}{50}\)
----
\(P(B \text{ and } (\text{not } A))\)
----
\(P(A \text{ or } B)\)
----
\(T\)
----
\(V\)
----
\(P(\text{not } (T \text{ or } V))\)
----
\(S\)
----
\(X\)
----
\(Y\)
----
\(S\)
----
\(X\)
----
\(Y\)
----
\(A\)
----
\(B\)
----
\[P(A \text{ and } B) = P(A) \times P(B)\]
----
\(E\)
----
\(T\)
----
\(E\)
----
\(E\)
----
\(T\)
----
\(P(A\text{ and }B) = P(A) \times P(B)\)
----
\(A\)
----
\(B\)
----
\(B\)
----
\(A\)
----
\(A\)
----
\(A\)
----
\(S\)
----
\[P(A) = \frac{n(A)}{n(S)}\]
----
\(B\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(A\text{ and }B\)
----
\(B\)
----
\[P(A\text{ if we know }B) = \frac{n(A\text{ and }B)}{n(B)}\]
----
\(P(A) = P(A\text{ if we know }B)\)
----
\(B\)
----
\(A\)
----
\[P(A\text{ and }B) = P(A) \times P(B)\]
----
\[\begin{align*}
  P(A) &= P(A\text{ and }B) \div P(B) \\
  &= \frac{n(A\text{ and }B)}{n(S)} \div \frac{n(B)}{n(S)} \\
  &= \frac{n(A\text{ and }B)}{n(B)} \\
  &= P(A\text{ if we know }B)
\end{align*}\]
----
\[\frac{P(A \text{ and } B)}{P(B)}\]
----
\(P(A\;|\;B)\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(B\)
----
\(P(A\;|\;B) = P(A)\)
----
\(P(B\;|\;A) = P(B)\)
----
\[P(\text{first ball red}) = \frac{5}{10} = \frac{1}{2}\]
----
\[P(\text{second ball blue}) = \frac{5}{10} = \frac{1}{2}\]
----
\(\frac{5}{10}\)
----
\(\frac{5}{10}\)
----
\(\frac{1}{4}\)
----
\[P(A\text{ and }B) = P(A) \times P(B)\]
----
\(P(\text{first ball red}) = \frac{1}{2}\)
----
\(P(\text{second ball blue}) = \frac{1}{2}\)
----
\(P(\text{first ball red and second ball blue}) = \frac{1}{4}\)
----
\(\frac{1}{4} = \frac{1}{2}\times\frac{1}{2}\)
----
\(\frac{5}{10}\)
----
\(\frac{4}{9}\)
----
\[\frac{5}{10}\times\frac{4}{9}=\frac{20}{90}\]
----
\(\frac{2}{9}\)
----
\(\frac{5}{10}\)
----
\(\frac{5}{9}\)
----
\[\frac{5}{10}\times\frac{5}{9}=\frac{25}{90}\]
----
\(\frac{5}{18}\)
----
\(\frac{5}{18}\)
----
\(\frac{2}{9}\)
----
\(\frac{2}{9}\)
----
\(\frac{5}{18}\)
----
\(\frac{5}{18}\)
----
\(\frac{2}{9}\)
----
\(\frac{2}{9}\)
----
\(\frac{5}{18}\)
----
\[P(\text{first ball red}) = \frac{2}{9} + \frac{5}{18} = \frac{1}{2}\]
----
\(\frac{5}{18}\)
----
\(\frac{2}{9}\)
----
\[P(\text{second ball blue}) = \frac{5}{18} + \frac{2}{9} = \frac{1}{2}\]
----
\(\frac{1}{2}\)
----
\(\frac{5}{18}\)
----
\[P(A\text{ and }B) = P(A) \times P(B)\]
----
\(P(\text{first ball red}) = \frac{1}{2}\)
----
\(P(\text{second ball blue}) = \frac{1}{2}\)
----
\(P(\text{first ball red and second ball blue}) = \frac{5}{18}\)
----
\(\frac{5}{18} \ne \frac{1}{2}\times\frac{1}{2}\)
----
\(P(A \text{ and } B) = 0\)
----
\(P(A \text{ and } B) = P(A) \times P(B)\)
----
\(X\)
----
\(Y\)
----
\((X \text{ and } Y) = \emptyset\)
----
\((X \text{ and } Y)\)
----
\(X\)
----
\(Y\)
----
\(P(X \text{ and } Y) = P(X) \times P(Y)\)
----
\[\begin{align}
		    P(X \text{ and } Y) &= \frac{3}{35} \\
		    P(X) &= \frac{14}{35} \\
		    P(Y) &= \frac{10}{35}
		    \end{align}\]
----
\(P(X) \times P(Y) = \frac{4}{35} \ne P(X \text{ and } Y)\)
----
\(X\)
----
\(Y\)
----
\(\frac{17}{30}\)
----
\(\frac{11}{30}\)
----
\(P(\text{black and brown}) = 0\)
----
\(P(\text{black and brown}) = 0 \ne P(\text{black}) \times P(\text{brown})\)
----
\(M\)
----
\(N\)
----
\(P(M \text{ and } N) \ne 0\)
----
\(A\)
----
\(B\)
----
\(P(A) \ne 0\)
----
\(P(B) \ne 0\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(B\)
----
\(P(A \text{ and } B) = P(A) \times P(B)\)
----
\(A\)
----
\(B\)
----
\(P(A \text{ and } B) = 0\)
----
\(P(A) \times P(B) = 0\)
----
\(P(A) \ne 0\)
----
\(P(B) \ne 0\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(B\)
----
\(S\)
----
\(P\)
----
\(M\)
----
\(F\)
----
\[S = \{1; 2; 3; \ldots; 30\}\]
----
\[P = \{2; 3; 5; 7; 11; 13; 17; 19; 23; 29\}\]
----
\[M = \{3; 6; 9; 12; 15; 18; 21; 24; 27; 30\}\]
----
\[F = \{1; 2; 3; 5; 6; 10; 15; 30\}\]
----
\(P\)
----
\(M\)
----
\(F\)
----
\(S\)
----
\(A \text{ and } B \text{ and } C\)
----
\(A \cap B \cap C\)
----
\(A \text{ or } B \text{ or } C\)
----
\(A \cup B \cup C\)
----
\(A \text{ and } B\)
----
\(A \cap B\)
----
\(A \text{ or } B\)
----
\(A \cup B\)
----
\(P(F)\)
----
\(P(G\text{ or }H)\)
----
\(P(F\text{ and }G)\)
----
\(F\)
----
\(G\)
----
\[\begin{align}
		  P(F) \times P(G)
		  &= \frac{9}{40} \times \frac{51}{120} \\
		  &= \frac{153}{1600}
		  \end{align}\]
----
\[P(F\text{ and }G) = \frac{1}{10}\]
----
\(P(F\text{ and }G) \ne P(F) \times P(G)\)
----
\[P\left(Z\text{ and not }(X\text{ or }Y)\right)\]
----
\(\left(Z\text{ and not }(X\text{ or }Y)\right)\)
----
\(Z\)
----
\(X\)
----
\(Y\)
----
\(P\left(Z\text{ and not }(X\text{ or }Y)\right) = \frac{4}{25}\)
----
\(30+36-16=50\)
----
\(79-50=29\)
----
\(79-41-8-16 = 14\)
----
\(A\)
----
\(B\)
----
\(C\)
----
\[P(A\text{ or }B\text{ or }C) = P(A) + P(B) + P(C)\]
----
\(A\)
----
\(B\)
----
\(C\)
----
\(P(A\text{ or }B\text{ or }C) = P(A) + P(B) + P(C)\)
----
\(\frac{1}{6}\)
----
\[P(A\text{ and }B) = P(A) \times P(B)\]
----
\(\frac{1}{3}\)
----
\(\frac{1}{6}\)
----
\(\frac{1}{5}\)
----
\(\frac{1}{5}\)
----
\(\frac{1}{3}\)
----
\(\frac{1}{6}\)
----
\[\begin{align}
  &P(\text{rain tomorrow and rain day after tomorrow}) \\
  =& \frac{1}{5} \times \frac{1}{3} \\
  =& \frac{1}{15}
\end{align}\]
----
\[\begin{align}
  &P(\text{not rain tomorrow and rain day after tomorrow}) \\
  =& \frac{4}{5} \times \frac{1}{6} \\
  =& \frac{2}{15}
\end{align}\]
----
\[\frac{1}{15} + \frac{2}{15} = \frac{1}{5}\]
----
\(H\)
----
\(T\)
----
\(\frac{1}{2}\)
----
\((H; T)\)
----
\(\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\)
----
\((H; H; H)\)
----
\(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}\)
----
\(\frac{1}{4}+\frac{1}{8}=\frac{3}{8}\)
----
\(\frac{1}{2}\)
----
\(\frac{1}{2}\)
----
\(\frac{3}{5}\)
----
\(\frac{1}{10}\)
----
\(\frac{2}{5}\)
----
\(\frac{9}{10}\)
----
\[\begin{align*}
  \frac{1}{2} \times \frac{3}{5} &= \frac{3}{10} \\
  \frac{1}{2} \times \frac{1}{10} &= \frac{1}{20}
\end{align*}\]
----
\(\frac{7}{20}\)
----
\(\frac{1}{6}\)
----
\(\frac{1}{6}\times\frac{1}{6} = \frac{1}{36}\)
----
\(\frac{7}{36}\)
----
\(2^4=16\)
----
\(\{H;H;H;T\}\)
----
\(\{H;H;T;H\}\)
----
\(\{H;T;H;H\}\)
----
\(\{T;H;H;H\}\)
----
\(\{H;H;H;H\}\)
----
\(\dfrac{5}{16}\)
----
\(\dfrac{1}{4}\)
----
\(\dfrac{5}{16}\)
----
\[\begin{align*}
  P(\text{negative if male})
  &= \frac{n(\text{negative and male})}{n(\text{male})} \\
  &= \frac{60}{90} \\
  &= \frac{2}{3}
\end{align*}\]
----
\[P(A\text{ and }B) = P(A) \times P(B)\]
----
\[P(\text{female and positive}) \ne P(\text{female}) \times P(\text{positive})\]
----
\(\frac{59}{65}\)
----
\(\frac{6}{9} = \frac{2}{3}\)
----
\(\frac{8}{23}\)
----
\(\frac{12}{23}\)
----
\(\frac{5}{23}\)
----
\(\frac{5}{23} \ne \frac{8}{23} \times \frac{12}{23}\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(A\)
----
\(B\)
----
\(B\)
----
\(n(A) = 50 - 15 = 35\)
----
\(n(\text{not } B) = 50 - 20 = 30\)
----
\(A\)
----
\(A\)
----
\(B\)
----
\(B\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(A\)
----
\(B\)
----
\(B\)
----
\(A \text{ or } B\)
----
\(A \text{ and } B\)
----
\((A \text{ and } B) = \emptyset\)
----
\(\text{not } A\)
----
\(P(A\text{ and }B) = P(A) \times P(B)\)
----
\(P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\)
----
\(P(A \text{ or } B) = P(A) + P(B)\)
----
\(P(\text{not } A) = 1 - P(A)\)
----
\(D\)
----
\(F\)
----
\(P(F)\)
----
\(D\)
----
\(F\)
----
\[P(D \text{ or } F) = P(D) + P(F)\]
----
\(A\)
----
\(B\)
----
\(A\)
----
\(A\)
----
\(A\)
----
\(B\)
----
\(B\)
----
\(B\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(B\)
----
\(X\)
----
\(Y\)
----
\(X\)
----
\(Y\)
----
\(P(H')\)
----
\(P(H \text{ or } J)\)
----
\(P(H' \text{ or } J')\)
----
\((H' \text{ or } J')\)
----
\(H'\)
----
\(J'\)
----
\(H' \text{ or } J'\)
----
\(P(H' \text{ or } J)\)
----
\((H' \text{ or } J)\)
----
\(H'\)
----
\(J\)
----
\(H' \text{ or } J\)
----
\(P(H' \text{ and } J')\)
----
\((H' \text{ and } J')\)
----
\(H'\)
----
\(J'\)
----
\(H' \text{ and } J'\)
----
\(D\)
----
\(E\)
----
\(F\)
----
\(P(\text{not } F)\)
----
\(P(F \text{ or } D)\)
----
\(P(\text{neither } E \text{ nor } F)\)
----
\(D\)
----
\(P(D \text{ and } E)\)
----
\(D\)
----
\(E\)
----
\(P(E \text{ and } F)\)
----
\(P(E \text{ and } D')\)
----
\(D\)
----
\(E\)
----
\(D' = E\)
----
\(A\)
----
\(B\)
----
\(C\)
----
\(A\)
----
\(B\)
----
\(C\)
----
\(A\)
----
\(C\)
----
\(B\)
----
\(C\)
----
\(n(A\text{ and }B\text{ and }C) = 7\)
----
\(A\)
----
\(C\)
----
\(n(A\text{ and }C) = 21\)
----
\(B\)
----
\(C\)
----
\(n(B\text{ and }C) = 18\)
----
\(C\)
----
\(n(C) = 40\)
----
\(n(A\text{ or }B\text{ or }C) = 68\)
----
\(n(A\text{ or }B) = 60\)
----
\[n(A\text{ and }B) = n(A) + n(B) - n(A\text{ or }B) = 40 + 35 - 60 = 15\]
----
\(A\)
----
\(B\)
----
\(C\)
----
\(G\)
----
\(H\)
----
\(P(H) = h\)
----
\(h\)
----
\(G\)
----
\(H\)
----
\(G\)
----
\(H\)
----
\(P(G \text{ and } H) = 0\)
----
\[P(G \text{ or } H) = P(G) + P(H) - P(G \text{ and } H)\]
----
\(P(G \text{ and } H) = 0\)
----
\(h\)
----
\(G\)
----
\(H\)
----
\(G\)
----
\(H\)
----
\(P(G \text{ and } H) = P(G) \times P(H)\)
----
\[P(G \text{ or } H) = P(G) + P(H) - P(G \text{ and } H)\]
----
\(P(G \text{ and } H) = P(G) \times P(H)\)
----
\[\begin{array}{r@{\;}l}
  P(\text{two orange balls})
  &= \left(\frac{10}{17}\times\frac{9}{16}\times\frac{7}{15}\right) + \left(\frac{10}{17}\times\frac{7}{16}\times\frac{9}{15}\right) + \left(\frac{7}{17}\times\frac{10}{16}\times\frac{9}{15}\right) \\[4pt]
  &= \frac{63}{136}
\end{array}\]
----
\(P(\text{two orange balls}) = \frac{63}{136}\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(B\)
----
\(A\)
----
\(B\)
----
\(P(\text{Multivitamin A}) = \frac{540}{960} = \frac{9}{16}\)
----
\(P(\text{Improvement in health}) = \frac{700}{960} = \frac{35}{48}\)
----
\(P(\text{Multivitamin A and Improvement in health}) = \frac{400}{960} = \frac{5}{12}\)
----
\(\frac{9}{16} \times \frac{35}{48} = \frac{105}{256} \ne \frac{5}{12}\)
----
\(A\)
----
\(B\)
----
\((0;0)\)
----
\[\overline{x} = \frac{\sum_{i=1}^n x_i}{n}\]
----
\(x_i\)
----
\(n\)
----
\(\overline{x}\)
----
\[14\ ;\ 17\ ;\ 45\ ;\ 20\ ;\ 19\ ;\ 36\ ;\ 7\ ;\ 30\ ;\ 8\]
----
\[7\ ;\ 8\ ;\ 14\ ;\ 17\ ;\ 19\ ;\ 20\ ;\ 30\ ;\ 36\ ;\ 45\]
----
\(45 - 7 = 38\)
----
\(30 - 14 = 16\)
----
\[3\ ;\ 6\ ;\ 7\ ;\ 7\ ;\ 9\ ;\ 11\ ;\ 12\ ;\ 16\ ;\ 31\ ;\ 35\ ;\ 40\ ;\ 45\]
----
\(\frac{31+35}{2}=33\)
----
\(x\)
----
\(y\)
----
\(x\)
----
\(y\)
----
\(\frac{2}{16} = \frac{1}{8}\)
----
\(\frac{5}{16}\)
----
\(\frac{9}{16}\)
----
\(\frac{1}{8}\)
----
\(\frac{5}{16}\)
----
\(\frac{9}{16}\)
----
\(x\)
----
\(86 + 84 = 170\)
----
\(86 - 83 = 3\)
----
\(y\)
----
\(y\)
----
\(10 < n \le 20\)
----
\(20 < n \le 30\)
----
\(30 < n \le 40\)
----
\(40 < n \le 50\)
----
\(50 < n \le 60\)
----
\(10 < n \le 20\)
----
\(20 < n \le 30\)
----
\(30 < n \le 40\)
----
\(40 < n \le 50\)
----
\(50 < n \le 60\)
----
\(y\)
----
\((10;0)\)
----
\((20;5)\)
----
\((30;12)\)
----
\(\frac{1}{4}\)
----
\(\frac{1}{2}\)
----
\(\frac{3}{4}\)
----
\((1; 3; 7; 8; 10)\)
----
\([20,30)\)
----
\([30,40)\)
----
\([40,50)\)
----
\([50,60)\)
----
\([60,70)\)
----
\([70,80)\)
----
\([80,90)\)
----
\([90,100)\)
----
\(x\)
----
\([-25;-15)\)
----
\([-25;-15)\)
----
\([-15;-5)\)
----
\([-5;5)\)
----
\([5;15)\)
----
\([15;25)\)
----
\([0;8)\)
----
\([8;16)\)
----
\([16;24)\)
----
\([24;32)\)
----
\([32;40)\)
----
\([40;48)\)
----
\([48;56)\)
----
\([56;64)\)
----
\([64;72)\)
----
\([72;80)\)
----
\(n\)
----
\(\{x_1; x_2; \ldots; x_n\}\)
----
\(\overline{x}\)
----
\[\sigma^2 = \frac{\sum_{i=1}^n \left(x_i - \overline{x}\right)^2}{n}\]
----
\(\sigma^2\)
----
\[\{44; 49; 52; 62; 53; 48; 54; 49; 46; 51\}\]
----
\[\overline{x} = \frac{\sum_{i=1}^n x_i}{n}\]
----
\[\sigma^2 = \frac{\sum_{i=1}^n \left(x_i - \overline{x}\right)^2}{n}\]
----
\(x_i\)
----
\(x_i - \overline{x}\)
----
\((x_i - \overline{x})^2\)
----
\(n\)
----
\(\{x_1; x_2; \ldots; x_n\}\)
----
\(\overline{x}\)
----
\[\sigma = \sqrt{\frac{\sum_{i=1}^n \left(x_i - \overline{x}\right)^2}{n}}\]
----
\(\overline{x}\)
----
\(i\)
----
\(x_i\)
----
\(x_i - \overline{x}\)
----
\((x_i - \overline{x})^2\)
----
\(\sum x_i = \ldots\)
----
\(\sum (x_i - \overline{x}) = \ldots\)
----
\(\sum (x_i - \overline{x})^2 = \ldots\)
----
\(x=\left\{1;2;3;4;5;6\right\}\)
----
\(n = 6\)
----
\[\sigma = \sqrt{\frac{\sum {\left(x-\overline{x}\right)}^{2}}{n}}\]
----
\(a\)
----
\(b\)
----
\(a = 33-b\)
----
\(b = 13\)
----
\(b = 20\)
----
\(a = 33-b\)
----
\(a = 20\)
----
\(a = 13\)
----
\(a\)
----
\(b\)
----
\(\sigma^2 = \frac{1}{n}\sum_{i=1}^n (x_i - \overline{x})^2\)
----
\(\sigma = \sqrt{\frac{1}{n}\sum_{i=1}^n (x_i - \overline{x})^2}\)
----
\(11 \frac{3}{8}\)
----
\(5 \frac{3}{8}\)
----
\(M\)
----
\(R\)
----
\(M\)
----
\(R\)
----
\((0;0)\)
----
\((0;1)\)
----
\((0;2)\)
----
\((0;3)\)
----
\((1;0)\)
----
\((1;1)\)
----
\((1;2)\)
----
\((1;3)\)
----
\((2;0)\)
----
\((2;1)\)
----
\((2;2)\)
----
\((2;3)\)
----
\((3;0)\)
----
\((3;1)\)
----
\((3;2)\)
----
\((3;3)\)
----
\((4;0)\)
----
\((4;1)\)
----
\((4;2)\)
----
\((4;3)\)
----
\((5;0)\)
----
\((5;1)\)
----
\((5;2)\)
----
\((5;3)\)
----
\(1(M) + 2(R) \leq 8\)
----
\(M + 2R > 8\)
----
\(M\)
----
\(R\)
----
\((0;0)\)
----
\((0;1)\)
----
\((0;2)\)
----
\((0;3)\)
----
\((1;0)\)
----
\((1;1)\)
----
\((1;2)\)
----
\((1;3)\)
----
\((2;0)\)
----
\((2;1)\)
----
\((2;2)\)
----
\((2;3)\)
----
\((3;0)\)
----
\((3;1)\)
----
\((3;2)\)
----
\(\cancel{(3;3)}\)
----
\((4;0)\)
----
\((4;1)\)
----
\((4;2)\)
----
\(\cancel{(4;3)}\)
----
\((5;0)\)
----
\((5;1)\)
----
\(\cancel{(5;2)}\)
----
\(\cancel{(5;3)}\)
----
\((5;3)\)
----
\(5 + 2(3) = 11\)
----
\(P\)
----
\(M\)
----
\(R\)
----
\(M\)
----
\(R\)
----
\(\cancel{(3;3)}\)
----
\(\cancel{(4;3)}\)
----
\(\cancel{(5;2)}\)
----
\(\cancel{(5;3)}\)
----
\(K\)
----
\(T\)
----
\(K\)
----
\(T\)
----
\((10;10)\)
----
\((10;15)\)
----
\((10;20)\)
----
\((10;25)\)
----
\((10;30)\)
----
\((15;10)\)
----
\((15;15)\)
----
\((15;20)\)
----
\((15;25)\)
----
\((15;30)\)
----
\((20;10)\)
----
\((20;15)\)
----
\((20;20)\)
----
\((20;25)\)
----
\((20;30)\)
----
\((25;10)\)
----
\((25;15)\)
----
\((25;20)\)
----
\((25;25)\)
----
\((25;30)\)
----
\((30;10)\)
----
\((30;15)\)
----
\((30;20)\)
----
\((30;25)\)
----
\((30;30)\)
----
\(K + T < 40\)
----
\(K\)
----
\(T\)
----
\(\cancel{(10;10)}\)
----
\(\cancel{(10;15)}\)
----
\(\cancel{(10;20)}\)
----
\(\cancel{(10;25)}\)
----
\((10;30)\)
----
\(\cancel{(15;10)}\)
----
\(\cancel{(15;15)}\)
----
\(\cancel{(15;20)}\)
----
\((15;25)\)
----
\((15;30)\)
----
\(\cancel{(20;10)}\)
----
\(\cancel{(20;15)}\)
----
\((20;20)\)
----
\((20;25)\)
----
\((20;30)\)
----
\(\cancel{(25;10)}\)
----
\((25;15)\)
----
\((25;20)\)
----
\((25;25)\)
----
\((25;30)\)
----
\((30;10)\)
----
\((30;15)\)
----
\((30;20)\)
----
\((30;25)\)
----
\((30;30)\)
----
\(K\)
----
\(T\)
----
\(K\)
----
\(T\)
----
\(\cancel{(10;10)}\)
----
\(\cancel{(10;15)}\)
----
\(\cancel{(10;20)}\)
----
\(\cancel{(10;25)}\)
----
\(\cancel{(15;10)}\)
----
\(\cancel{(15;15)}\)
----
\(\cancel{(15;20)}\)
----
\(\cancel{(20;10)}\)
----
\(\cancel{(20;15)}\)
----
\(\cancel{(25;10)}\)
----
\(M\)
----
\(R\)
----
\(M\)
----
\(R\)
----
\[\begin{align}
								\text{number of Mountees: } 0 \leq M &\leq  5 \\
								\text{number of Roadees: } 0 \leq R &\leq  3 \\
								\text{total number of technicians: } M + 2R &\leq  8
									\end{align}\]
----
\(P\)
----
\(M\)
----
\(R\)
----
\(M\)
----
\(R\)
----
\(M + 2R \leq 8\)
----
\(R\)
----
\(y\)
----
\(R \leq -\frac{1}{2}M + 4\)
----
\(R \leq -\frac{1}{2}M + 4\)
----
\(k\)
----
\(t\)
----
\(k, t \in \mathbb{Z}\)
----
\[\begin{align}
		\text{number of kettles: } k &\geq  10 \\
		\text{number of toasters: } t &\geq  10 \\
		\text{total number of prizes: } k + t &\geq  40
		\end{align}\]
----
\(t\)
----
\[t \geq -k + 40\]
----
\(C\)
----
\[C = 120k + 100t\]
----
\(t\)
----
\[t = -\frac{6}{5}k + \frac{C}{100}\]
----
\(-\frac{6}{5}\)
----
\(t\)
----
\(\frac{C}{100}\)
----
\(C\)
----
\(t\)
----
\(A(10;30)\)
----
\(A\)
----
\(A\)
----
\(C = 120k + 100t\)
----
\(t\)
----
\(x\)
----
\(y\)
----
\[\begin{aligned}
			x & \leq 10 \\ 
			y & \leq 10
		\end{aligned}\]
----
\[\begin{aligned}
			x & \geq 4
		\end{aligned}\]
----
\[\begin{aligned}
			2x + 3y & \leq 30 
		\end{aligned}\]
----
\(x \leq 10; \quad y  \leq 10; \quad  x  \geq 4; \quad 2x + 3y \leq 30\)
----
\(T\)
----
\(T\)
----
\[\begin{aligned}
				T = & (\text{5 marks}) \times (\text{algebra questions answered}) + \\
				& (\text{10 marks}) \times (\text{geometry questions answered}) \\
				T = & 5x + 10y
				\end{aligned}\]
----
\(T = 5x + 10y\)
----
\(x\)
----
\(y\)
----
\(I\)
----
\(I = 50x + 30y\)
----
\(x\)
----
\(y\)
----
\(x\)
----
\(y\)
----
\(x\)
----
\(y\)
----
\[\begin{aligned}
			y & \geq (0.5) x
			\end{aligned}\]
----
\[\begin{aligned}
			x + y & \leq 500
			\end{aligned}\]
----
\(P\)
----
\(x\)
----
\(y\)
----
\(x\)
----
\(y\)
----
\(0 \leq x \leq 150\)
----
\[x + y \leq 200\]
----
\[\begin{aligned}
	x & \geq 40 \\
	y & \geq 10 
	\end{aligned}\]
----
\(P\)
----
\(x\)
----
\(y\)
----
\(x\)
----
\(y\)
----
\[\begin{aligned}
				x + y & \leq 5
				\end{aligned}\]
----
\(0;5\)
----
\(5;0\)
----