## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 7 - Rational Functions - 7.3 Multiplying and Dividing Rational Expressions - 7.3 Exercises: 47

#### Answer

$= \frac{(3x-2)}{(4x-3)}$

#### Work Step by Step

$\frac{6x^{2}+x-35}{10x^{2}+39x+35} \div \frac{12x^{2}-37x+21}{15x^{2}+11x-14}$ $= \frac{6x^{2}+x-35}{10x^{2}+39x+35} \times \frac{15x^{2}+11x-14}{12x^{2}-37x+21}$ $= \frac{6x^{2}+15x-14x-35}{10x^{2}+25x+14x+35} \times \frac{15x^{2}+21x-10x-14}{12x^{2}-28x-9x+21}$ $= \frac{3x(2x+5)-7(2x+5)}{5x(2x+5)+7(2x+5)} \times \frac{3x(5x+7)-2(5x+7)}{4x(3x-7)-3(3x-7)}$ $= \frac{(3x-7)(2x+5)}{(5x+7)(2x+5)} \times \frac{(3x-2)(5x+7)}{(4x-3)(3x-7)}$ $= \frac{(3x-7)}{(5x+7)} \times \frac{(3x-2)(5x+7)}{(4x-3)(3x-7)}$ $= \frac{1}{(5x+7)} \times \frac{(3x-2)(5x+7)}{(4x-3)}$ $= \frac{1}{1} \times \frac{(3x-2)}{(4x-3)}$ $=1 \times \frac{(3x-2)}{(4x-3)}$ $= \frac{(3x-2)}{(4x-3)}$

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