College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.2 - Page 119: 54



Work Step by Step

We may substitute the expressions involving x into the equation $y_{1}-y_{2}=-4$ and solve: $(\frac{x+1}{4})-(\frac{x-2}{3})=-4$ $(\frac{3}{3})(\frac{x+1}{4})-(\frac{4}{4})(\frac{x-2}{3})=-4$ $(\frac{3x+3}{12})-(\frac{4x-8}{12})=-4$ $\frac{(3x+3)-(4x-8)}{12}=-4$ $\frac{3x+3-4x+8}{12}=-4$ $\frac{-x+11}{12}=-4$ $(12)(\frac{-x+11}{12})=(12)(-4)$ $-x+11=-48$ $-x=-59$ $x=59$ Because we are dealing with a linear equation in one variable, there is only one answer.
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