Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 6 - Rational Expressions and Equations - 6.2 Multiplication and Division - 6.2 Exercise Set - Page 385: 69

Answer

$$\frac{(w-7)(w-8)}{(2w-7)(3w+1)}$$

Work Step by Step

Recall, one never divides by a fraction. Rather, if we want to divide by a fraction, we multiply by the reciprocal of the fraction. Recall, the reciprocal of a fraction is that fraction with the numerator and the denominator switched. Thus, we find: $$ \frac{\left(w^2-14w+49\right)}{2w^2-3w-14}\times \frac{w^2-6w-16}{\left(3w^2-20w-7\right)}\\ \frac{\left(w^2-14w+49\right)\left(w^2-6w-16\right)}{\left(2w^2-3w-14\right)\left(3w^2-20w-7\right)}\\ \frac{\left(w-7\right)\left(w-8\right)}{\left(2w-7\right)\left(3w+1\right)}\\ \frac{\left(w-7\right)\left(w-8\right)}{6w^2-19w-7}\\ \frac{w^2-15w+56}{6w^2-19w-7}\\ \frac{(w-7)(w-8)}{(2w-7)(3w+1)}$$
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