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 1.6 - Page 181: 139

Answer

$x=14$

Work Step by Step

$\frac{x+3}{4} = \frac{x-2}{3} + \frac{1}{4}$ Taking LCD, $\frac{x+3}{4} = \frac{4(x-2)+3}{12} $ Multiply both sides by $12$ $12(\frac{x+3}{4}) = 12(\frac{4(x-2)+3}{12} )$ $3(x+3)= 4(x-2) + 3$ $3x+9=4x-8+3$ $3x+9=4x-5$ $4x-3x = 9+5$ $x = 14$
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