Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 5 - Exponents and Polynomials - Chapters 1-5 Cumulative Review Problem Set: 13

Answer

$-\frac{2}{13}$

Work Step by Step

We substitute the given values into the equation to obtain: $\frac{3a-2b-4a+7b}{-a-3a+b-2b}$ $a=-1$ $b=-\frac{1}{3}$ $\frac{3(-1)-2(-\frac{1}{3})-4(-1)+7(-\frac{1}{3})}{-(-1)-3(-1)+(-\frac{1}{3})-2(-\frac{1}{3})}=\frac{-3+\frac{2}{3}+4-\frac{7}{3}}{1+3-\frac{1}{3}+\frac{2}{3}}=\frac{-\frac{9}{3}+\frac{2}{3}+\frac{12}{3}-\frac{7}{3}}{\frac{3}{3}+\frac{9}{3}-\frac{1}{3}+\frac{2}{3}}=\frac{-\frac{2}{3}}{\frac{13}{3}}=-\frac{2}{3}\times\frac{3}{13}=-\frac{2}{13}$
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