# Rational Expression Arithmetic

1. Simplify:

1. $\displaystyle{\frac{3x^2 - 12}{x^2 + 4x + 4} \cdot \frac{x+2}{x-2}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{3(x^2-4)}{(x+2)^2} \cdot \frac{x+2}{x-2}}\\\\ &=& \displaystyle{\frac{3(x-2)(x+2)}{(x+2)^2} \cdot \frac{x+2}{x-2}}\\\\ &=& \displaystyle{\frac{(x-2)(x+2)^2}{(x-2)(x+2)^2} \cdot 3}\\\\ &=& \fbox{$3$} \end{array}}$

2. $\displaystyle{\frac{x^2+3xy}{2x^3-x^2y} \cdot \frac{4x^2-y^2}{x^2+6xy+9y^2}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{x(x+3y)}{x^2(2x-y)} \cdot \frac{(2x+y)(2x-y)}{(x+3y)^2}}\\\\ &=& \displaystyle{\frac{x(x+3y)(2x-y)}{x(x+3y)(2x-y)} \cdot \frac{2x+y}{x(x+3y)}}\\\\ &=& \fbox{$\displaystyle{\frac{2x+y}{x(x+3y)}}$} \end{array}}$

3. $\displaystyle{\frac{x^2 - 4x - 21}{x^2 -9x + 18} \div \frac{x^2 - 49}{x-3}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{x^2 - 4x - 21}{x^2 -9x + 18} \cdot \frac{x-3}{x^2 - 49}}\\\\ &=& \displaystyle{\frac{(x-7)(x+3)}{(x-6)(x-3)} \cdot \frac{x-3}{(x-7)(x+7)}}\\\\ &=& \displaystyle{\frac{(x-7)(x-3)}{(x-7)(x-3)} \cdot \frac{x+3}{(x-6)(x+7)}}\\\\ &=& \fbox{$\displaystyle{\frac{x+3}{x^2+x-42}}$} \end{array}}$

4. $\displaystyle{\frac{x^2 + 3xy}{2x^3 - x^2 y} \, \div \, \frac{x^2 + 6xy + 9y^2}{4x^2 - y^2}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{x^2 + 3xy}{2x^3 - x^2 y} \, \cdot \, \frac{4x^2 - y^2}{x^2 + 6xy + 9y^2}}\\\\ &=& \displaystyle{\frac{x(x+3y)}{x^2(2x-y)} \, \cdot \, \frac{(2x+y)(2x-y)}{(x+3y)^2}}\\\\ &=& \displaystyle{\frac{x(2x-y)(x+3y)}{x(2x-y)(x+3y)} \cdot \frac{2x+y}{x(x+3y)}}\\\\ &=& \fbox{$\displaystyle{\frac{2x+y}{x(x+3y)}}$} \end{array}}$

5. $\displaystyle{\frac{4b}{b^2+6b+5} + \frac{2b}{b^2-1}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{ \frac{4b}{(b+1)(b+5)} + \frac{2b}{(b+1)(b-1)}}\\\\ &=& \displaystyle{ \frac{4b(b-1)}{(b+1)(b+5)(b-1)} + \frac{2b(b+5)}{(b+1)(b-1)(b+5)}}\\\\ &=& \displaystyle{ \frac{(4b^2 - 4b) + (2b^2+10b)}{(b+1)(b+5)(b-1)}}\\\\ &=& \displaystyle{ \frac{6b^2 + 6b}{(b+1)(b+5)(b-1)}}\\\\ &=& \displaystyle{ \frac{6b(b+1)}{(b+1)(b+5)(b-1)}}\\\\ &=& \displaystyle{ \frac{b+1}{b+1} \cdot \frac{6b}{(b+5)(b-1)}}\\\\ &=& \fbox{$\displaystyle{\frac{6b}{b^2+4b-5}}$} \end{array}}$

6. $\displaystyle{\frac{2}{2x^2 -5x -3} \, - \, \frac{1}{3x^2 -10x + 3}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{2}{(2x+1)(x-3)} \, - \, \frac{1}{(3x-1)(x-3)}}\\\\ &=& \displaystyle{\frac{2(3x-1)}{(2x+1)(x-3)(3x-1)} \, - \, \frac{2x+1}{(3x-1)(x-3)(2x+1)}}\\\\ &=& \displaystyle{\frac{(6x-2) - (2x+1)}{(x-3)(2x+1)(3x-1)}}\\\\ &=& \fbox{$\displaystyle{\frac{4x-3}{(x-3)(2x+1)(3x-1)}}$} \end{array}}$

7. $\displaystyle{\frac{x}{x^2+9x+20} - \frac{4}{x^2+7x+12}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{x}{(x+4)(x+5)} - \frac{4}{(x+3)(x+4)}}\\\\ &=& \displaystyle{\frac{x(x+3)}{(x+4)(x+5)(x+3)} - \frac{4(x+5)}{(x+3)(x+4)(x+5)}}\\\\ &=& \displaystyle{\frac{(x^2+3x) - (4x+20)}{(x+3)(x+4)(x+5)}}\\\\ &=& \displaystyle{\frac{x^2-x-20}{(x+3)(x+4)(x+5)}}\\\\ &=& \displaystyle{\frac{(x-5)(x+4)}{(x+3)(x+4)(x+5)}}\\\\ &=& \displaystyle{\frac{x+4}{x+4} \cdot \frac{x-5}{(x+3)(x+5)}}\\\\ &=& \fbox{$\displaystyle{\frac{x-5}{(x+3)(x+5)}}$} \end{array}}$

8. $\displaystyle{\frac{x}{x^2-x-6} - \frac{1}{x+2} + \frac{2}{3-x}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{x}{(x-3)(x+2)} - \frac{1}{x+2} - \frac{2}{x-3}}\\\\ &=& \displaystyle{\frac{x}{(x-3)(x+2)} - \frac{x-3}{(x+2)(x-3)} - \frac{2(x+2)}{(x-3)(x+2)}}\\\\ &=& \displaystyle{\frac{x - (x-3) - 2(x+2)}{(x-3)(x+2)}}\\\\ &=& \displaystyle{\frac{x - x + 3 - 2x - 4}{(x-3)(x+2)}}\\\\ &=& \displaystyle{\frac{-2x-1}{(x-3)(x+2)}}\\\\ &=& \fbox{$\displaystyle{\frac{-2x-1}{x^2-x-6}}$} \end{array}}$

9. $\displaystyle{\frac{1 - \displaystyle{\frac{6}{x} + \frac{5}{x^2}}}{\displaystyle{\frac{1}{x} - \frac{5}{x^2}}}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{\displaystyle{ \left( 1 - \frac{6}{x} + \frac{5}{x^2} \right)}}{\displaystyle{\left( \frac{1}{x} - \frac{5}{x^2} \right)}} \cdot \frac{x^2}{x^2}}\\\\ &=& \displaystyle{\frac{x^2 - 6x + 5}{x-5}}\\\\ &=& \displaystyle{\frac{(x-5)(x-1)}{x-5}}\\\\ &=& \displaystyle{\frac{x-5}{x-5} \cdot (x-1)}\\\\ &=& \fbox{$\displaystyle{x-1}$} \end{array}}$

10. $\displaystyle{\frac{\displaystyle{\frac{x}{1+x}} + \displaystyle{\frac{2}{x}}}{\displaystyle{\frac{1}{x}} - \displaystyle{\frac{1}{x+1}}}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{ \displaystyle{\left( \frac{x}{1+x} + \frac{2}{x} \right) }}{\displaystyle{ \left( \frac{1}{x} - \frac{1}{x+1} \right)}} \cdot \frac{x(x+1)}{x(x+1)}}\\\\ &=& \displaystyle{\frac{x^2 + 2(x+1)}{(x+1) - x}}\\\\ &=& \displaystyle{\frac{x^2 + 2x + 2}{1}}\\\\ &=& \fbox{$\displaystyle{x^2+2x+2}$} \end{array}}$

11. $\displaystyle{\frac{\displaystyle{\frac{x^2}{y}} + \displaystyle{\frac{y^2}{x}}}{y^2 - xy + x^2}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{\displaystyle{\left( \frac{x^2}{y} + \frac{y^2}{x} \right)}}{\displaystyle{y^2 - xy + x^2}} \cdot \frac{xy}{xy}}\\\\ &=& \displaystyle{\frac{x^3 + y^3}{(y^2 - xy + x^2)xy}}\\\\ &=& \displaystyle{\frac{(x+y)(x^2-xy+y^2)}{xy(x^2-xy+y^2)}}\\\\ &=& \displaystyle{\frac{x+y}{xy} \cdot \frac{x^2-xy+y^2}{x^2-xy+y^2} }\\\\ &=& \fbox{$\displaystyle{\frac{x+y}{xy}}$} \end{array}}$

12. $\displaystyle{\frac{\displaystyle{\frac{5}{m}} - \displaystyle{\frac{2}{m+1}}}{\displaystyle{\frac{3}{m+1}} + \displaystyle{\frac{1}{m}}}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{\displaystyle{\left( \frac{5}{m} - \frac{2}{m+1} \right)}}{\displaystyle{\left( \frac{3}{m+1} + \frac{1}{m} \right)}} \cdot \frac{m(m+1)}{m(m+1)}}\\\\ &=& \displaystyle{\frac{5(m+1) - 2m}{3m + (m+1)}}\\\\ &=& \displaystyle{\frac{5m + 5 - 2m}{3m + m + 1}}\\\\ &=& \fbox{$\displaystyle{\frac{3m+5}{4m+1}}$} \end{array}}$

13. $\displaystyle{\frac{b-a^{-1}}{a - b^{-1}}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{(b-a^{-1})}{(a - b^{-1})} \cdot \frac{ab}{ab}}\\\\ &=& \displaystyle{\frac{ab^2 - b}{a^2b - a}}\\\\ &=& \displaystyle{\frac{b(ab - 1)}{a(ab - 1)}}\\\\ &=& \displaystyle{\frac{b}{a} \cdot \frac{ab-1}{ab-1}}\\\\ &=& \fbox{$\displaystyle{\frac{b}{a}}$} \end{array}}$

14. $\displaystyle{\frac{x^{-2} - y^{-2}}{x^{-1} + \displaystyle{\frac{1}{y}}}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{(x^{-2} - y^{-2})}{\displaystyle{\left( x^{-1} + \frac{1}{y} \right)}} \cdot \frac{x^2 y^2}{x^2 y^2}}\\\\ &=& \displaystyle{\frac{y^2 - x^2}{xy^2 + x^2y}}\\\\ &=& \displaystyle{\frac{(y+x)(y-x)}{xy(y+x)}}\\\\ &=& \displaystyle{\frac{y+x}{y+x} \cdot \frac{y-x}{xy}}\\\\ &=& \fbox{$\displaystyle{\frac{y-x}{xy}}$} \end{array}}$

15. $\displaystyle{\frac{4x^3(x^2+4)^2 - x^4 \cdot 2 (x^2 + 4) \cdot 2x}{[(x^2+4)^2]^2}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{4x^3(x^2+4)[(x^2+4)-x^2]}{(x^2+4)^4}}\\\\ &=& \displaystyle{\frac{x^2+4}{x^2+4} \cdot \frac{4x^3(x^2 + 4 - x^2)}{(x^2+4)^3}}\\\\ &=& \displaystyle{\frac{4x^3(4)}{(x^2+4)^3}}\\\\ &=& \fbox{$\displaystyle{\frac{16x^3}{(x^2+4)^3}}$} \end{array}}$

2. Peform the indicated operation and simplify:

1. $\displaystyle{\frac{x^3-y^3}{x^2+xy+y^2} \cdot \frac{x+y}{x^2-y^2}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{(x-y)(x^2+xy+y^2)}{x^2+xy+y^2} \cdot \frac{x+y}{(x+y)(x-y)}}\\\\ &=& \displaystyle{\frac{(x-y)(x^2+xy+y^2)(x+y)}{(x-y)(x^2+xy+y^2)(x+y)}}\\\\ &=& \fbox{$\displaystyle{1}$} \end{array}}$

2. $\displaystyle{\frac{2x}{x-y} + \frac{3y}{y-x}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{2x}{x-y} - \frac{3y}{x-y}}\\\\ &=& \fbox{$\displaystyle{\frac{2x-3y}{x-y}}$} \end{array}}$

3. $\displaystyle{\frac{\displaystyle{\frac{x-2}{x}+\frac{x}{x+2}}}{\displaystyle{\frac{x-2}{x+2}+\frac{1}{x}}}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{\displaystyle{\left( \frac{x-2}{x}+\frac{x}{x+2} \right)}}{\displaystyle{\left( \frac{x-2}{x+2}+\frac{1}{x} \right)}} \cdot \frac{x(x+2)}{x(x+2)}}\\\\ &=& \displaystyle{\frac{(x-2)(x+2) + x^2}{x(x-2)+(x+2)}}\\\\ &=& \displaystyle{\frac{(x^2-4) + x^2}{x^2-2x+x+2}}\\\\ &=& \displaystyle{\frac{2x^2-4}{x^2-2x+x+2}}\\\\ &=& \fbox{$\displaystyle{\frac{2x^2-4}{x^2-x+2}}$} \end{array}}$

4. $\displaystyle{\frac{\displaystyle{\frac{x}{y}+\frac{y}{x}}}{\displaystyle{\frac{x^2}{y^2}-\frac{y^2}{x^2}}}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{\displaystyle{\left( \frac{x}{y}+\frac{y}{x} \right)}}{\displaystyle{\left( \frac{x^2}{y^2}-\frac{y^2}{x^2} \right)}} \cdot \frac{x^2 y^2}{x^2 y^2}}\\\\ &=& \displaystyle{\frac{x^3 y+xy^3}{x^4 - y^4}}\\\\ &=& \displaystyle{\frac{xy(x^2+y^2)}{(x^2+y^2)(x^2-y^2)}}\\\\ &=& \displaystyle{\frac{x^2+y^2}{x^2+y^2} \cdot \frac{xy}{x^2-y^2}}\\\\ &=& \fbox{$\displaystyle{\frac{xy}{x^2-y^2}}$} \end{array}}$

5. $\displaystyle{\frac{x^2-9}{2x+3} \div \frac{x+3}{2x^2-3x-9}}$   $\ans{ \begin{array}{rcl} &=& \displaystyle{\frac{x^2-9}{2x+3} \cdot \frac{2x^2-3x-9}{x+3}}\\\\ &=& \displaystyle{\frac{(x+3)(x-3)}{2x+3} \cdot \frac{(2x+3)(x-3)}{x+3}}\\\\ &=& \displaystyle{\frac{(x+3)(2x+3)}{(x+3)(2x+3)} \cdot (x-3)^2}\\\\ &=& \fbox{$\displaystyle{(x-3)^2}$} \end{array}}$