# Multiply, Divide and Simplify Rational Expressions - Grade 11 Math Questions With Detailed Solutions

Detailed solutions to questions in How to multiply, divide and simplify rational expressions are presented.

 Rules of multiplication, division and simplification of rational expressions? We multiply two rational expressions by multiplying their numerators and denominators as follows: 1) $\dfrac{A}{B} \cdot \dfrac{C}{D} = \dfrac{A \cdot C}{B \cdot D}$ We divide two rational expressions by multiplying the first by the reciprocal of the second as follows: 2) $\dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{A}{B} \cdot \dfrac{D}{C} = \dfrac{A \cdot D}{B \cdot C} \;\; \text{or} \;\; \dfrac{ \dfrac{A}{B} }{ \dfrac{C}{D}} = \dfrac{A}{B} \cdot \dfrac{D}{C} = \dfrac{A \cdot D}{B \cdot C}$ Multiply and/or divide and simplify the given rational expressions. a) $\dfrac{-3}{2} \div \dfrac{6x-9}{2x - 3}$ Solution: The division of two rational expressions is done by multiplying the first by the reciprocal of the second as follows (see divison rule above). Hence $\dfrac{-3}{2} \div \dfrac{6x-9}{2x-3} = \dfrac{-3}{2} \cdot \dfrac{2x-3} {6x-9}$ Multiply numerators and denominators (multiplication rule). $= \dfrac{-3(2x-3)}{2(6x-9)}$ Factor the terms 6x-9 included in the denominator as follows: 6x - 9 = 3(2x - 3) and use the factored form in the rational expression $= \dfrac{-3(2x-3)}{2\cdot3(2x-3)}$ simplify $= \dfrac{-{\colorcancel{red}{3}}{\colorcancel{red}{(2x-3)}}}{2 \cdot {\colorcancel{red}{3}}{\colorcancel{red}{(2x-3)}}} = -\dfrac{1}{2} \; \; \text{for} \; \; x \ne 3/2$ b) $\dfrac{2x-5}{2x+2} \cdot \dfrac{10x+10}{4x-10}$ Solution: Apply the multiplication rule (see above) $\dfrac{2x-5}{2x+2} \cdot \dfrac{10x+10}{4x-10} = \dfrac{(2x-5)(10x+10)}{(2x+2)(4x-10)}$ Factor terms in the numerator and the denominator: 10 x + 10 = 10(x + 1) ; 2 x + 2 = 2(x + 1) ; 4 x - 10 = 2(2x - 5) and use in factored form $= \dfrac{(2x-5) \cdot 10(x + 1)}{2(x+1)\cdot2(2x-5)}$ Simplify if possible $= \dfrac{{\colorcancel{red}{(2x-5)}} \cdot 10{\colorcancel{red}{(x+1)}}}{2{\colorcancel{red}{(x+1)}}\cdot 2{\colorcancel{red}{(2x-5)}}} = \dfrac{10}{4} = \dfrac{2 \cdot 5}{2 \cdot 2} = \dfrac{5}{2} \; \; \text{for} \; \; x \ne -1 \; \; \text{and} \; \; x \ne 5/2$ c) $\dfrac { \dfrac{2x^2-7x-15}{x^2+3x-4} }{ \dfrac{x^2+x-30}{x^2-1} }$ Solution: The division of two rational expressions is done by multiplying the first rational expression by the reciprocal of the second rational expression as follows (see divison rule above). Hence $\dfrac { \dfrac{2x^2-7x-15}{x^2+3x-4} }{ \dfrac{x^2+x-30}{x^2-1} } = \dfrac{2x^2-7x-15}{x^2+3x-4} \cdot \dfrac{x^2-1}{x^2+x-30}$ Multiply numerators and denominators (multiplication rule) but do not expand as we might be able to simplifty. $= \dfrac{(2x^2-7x-15)(x^2-1)}{(x^2+3x-4)(x^2+x-30)}$ factor terms in the numerator and denominator if possible. 2 x 2 - 7 x - 15 = (2x + 3)(x - 5) ; x 2 - 1 = (x - 1)(x + 1) x 2 + 3 x - 4 = (x + 4)(x - 1) ; x 2 + x - 30 = (x + 6)(x - 5) and use in rational expression $= \dfrac{(2x + 3)(x - 5)(x - 1)(x + 1)}{(x + 4)(x - 1)(x + 6)(x - 5)}$ and simplify. $= \dfrac{(2x + 3){\colorcancel{red}{(x-5)}}{\colorcancel{red}{(x-1)}}(x + 1)}{(x + 4){\colorcancel{red}{(x-1)}}(x + 6){\colorcancel{red}{(x-5)}}} = \dfrac{(2x + 3)(x + 1)}{(x + 4)(x + 6)} \; \; \text{for} \; \; x \ne 1 \text{} \;\; x \ne 5$ d) $(\dfrac{x-1}{x+2} \cdot \dfrac{x^2-4}{x^2-1}) \div \dfrac{x-2}{x+5}$ Solution: We have the multiplication of two rational expressions inside the parentheses and we then apply mutliplication rule. We also have a division by a rational expression which is done by multiplying by the reciprocal. Hence $(\dfrac{x-1}{x+2} \cdot \dfrac{x^2-4}{x^2-1}) \div \dfrac{x-2}{x+5} = \dfrac{(x-1)(x^2-4)}{(x+2)(x^2-1)} \cdot \dfrac{x+5}{x-2}$ Multiply numerators and denominators (multiplication rule) but do not expand as we might be able to simplifty. $= \dfrac{(x-1)(x^2-4)(x+5)}{(x+2)(x^2-1)(x-2)}$ Factor terms in numerator and denominator if possible and use in rational expression x 2 - 4 = (x - 2)(x + 2) and x 2 - 1 = (x - 1)(x + 1) $= \dfrac{(x-1)(x-2)(x+2)(x+5)}{(x+2)(x-1)(x+1)(x-2)} =$ Simplify $\dfrac{{\colorcancel{red}{(x-1)}}{\colorcancel{red}{(x-2)}}{\colorcancel{red}{(x+2)}}(x+5)}{{\colorcancel{red}{(x+2)}}{\colorcancel{red}{(x-1)}}(x+1){\colorcancel{red}{(x-2)}}} = \dfrac{x+5}{x+1} \; \; \text{for} \; \; x \ne - 2 \; \; x \ne 1 \; \; \text{and} \; \; x \ne 2$ e) $\dfrac{ \dfrac{x^3-27}{x+3} }{ \dfrac{x-3}{(x+3)^2}}$ Solution: The division of two rational expressions is done by multiplying the first rational expression by the reciprocal of the second rational expression as follows (see divison rule above). Hence $\dfrac{ \dfrac{x^3-27}{x+3} }{ \dfrac{x-3}{(x+3)^2}} = \dfrac{x^3-27}{x+3} \cdot \dfrac{(x+3)^2}{x-3}$ Multiply numerators and denominators (multiplication rule) but do not expand. $= \dfrac{(x^3-27)(x+3)^2}{(x+3)(x-3)}$ factor term x 3 - 27 in the numerator and use it. x 3 - 27 = (x - 3)(x 2 + 3 x + 9) and use in rational expression $= \dfrac{(x - 3)(x^2 + 3 x + 9)(x+3)^2}{(x+3)(x-3)}$ and simplify. $= \dfrac{{\colorcancel{red}{(x-3)}}(x^2 + 3 x + 9){\colorcancel{red}{(x+3)}(x+3)}}{{\colorcancel{red}{(x+3)}}{\colorcancel{red}{(x-3)}}} = (x +3)(x^2 + 3 x + 9) \; \; \text{for} \; \; x \ne - 3 \; \; \text{and} \; \; x \ne 3$ f) $\dfrac{2y-x}{4x^2-9y^2} \cdot \dfrac{4x+6y}{y-\dfrac{1}{2}x}$ Solution: Apply the multiplication rule (see above) $\dfrac{2y-x}{4x^2-9y^2} \cdot \dfrac{4x+6y}{y-\dfrac{1}{2}x} = \dfrac{(2y-x)(4x+6y)}{(4x^2-9y^2)(y-\dfrac{1}{2}x)}$ Factor terms in the numerator and the denominator: 2 y - x = 2 (y - (1/2) x) ; 4 x + 6 y = 2(2 x + 3) ; 4 x 2 - 9 y 2 = (2x - 3y)(2x + 3y) and use in factored form $= \dfrac{2(y-\dfrac{1}{2} x) \cdot 2(2x+3y)}{(2x-3y)(2x+3y)(y-\dfrac{1}{2}x)}$ Simplify if possible $= \dfrac{ 2{\colorcancel{red}{(y-\dfrac{1}{2} x)}} \cdot 2{\colorcancel{red}{(2x+3y)}} } {(2x-3y){\colorcancel{red}{(2x+3y)}}{\colorcancel{red}{(y-\dfrac{1}{2} x)}}} =\dfrac{4}{2x-3y} \; \; \text{for} \; \; x \ne 2y \; \; \text{and} \; \; x \ne -3y/2$

Updated: 20 January 2017 (A Dendane)