College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Review Exercises - Page 145: 8


$\displaystyle x=-\frac{27}{13}$

Work Step by Step

We solve: $\displaystyle \frac{1-3x}{4}=\frac{x+6}{3}+\frac{1}{2}$ $12*(\displaystyle \frac{1-3x}{4})=(\frac{x+6}{3}+\frac{1}{2})(12)$ $\frac{12}{4}(1-3x)=\frac{12}{3}(x+6)+\frac{12}{2}$ $3(1-3x)=4(x+6)+6$ $3-9x=4x+24+6$ $-13x=27$ $x=-\frac{27}{13}$
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