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: 30


$x= \frac{25}{7}$

Work Step by Step

$\frac{3x}{5}-\frac{x-3}{2} = \frac{x+2}{3}$ Take LCD at the left hand side of the expression $\frac{6x-5(x-3)}{10} = \frac{x+2}{3}$ $\frac{6x-5x+15}{10} = \frac{x+2}{3}$ $\frac{x+15}{10} = \frac{x+2}{3}$ Multiply both sides by 30 to clear fractional part. $30(\frac{x+15}{10}) = 30(\frac{x+2}{3})$ $3(x+15) = 10(x+2)$ $3x+45 = 10x+20$ $45-20 = 10x-3x$ $25 = 7x$ $7x = 25$ $x= \frac{25}{7}$
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