Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 1-3 - Cumulative Review - Page 228: 38



Work Step by Step

Multiply both sides by the LCD (which is 28) to obtain: $\require{cancel} 28 \left(\dfrac{2+5x}{4}\right)=28\left(\dfrac{11}{28}+\dfrac{8x+3}{7}\right) \\\cancel{28}7 \left(\dfrac{2+5x}{\cancel{4}}\right)=28\left(\dfrac{11}{28}\right)+28\left(\dfrac{8x+3}{7}\right) \\7(2+5x)=\cancel{28}\left(\dfrac{11}{\cancel{28}}\right)+\cancel{28}4\left(\dfrac{8x+3}{\cancel{7}}\right) \\7(2+5x)=11+4(8x+3)$ Distribute $7$ on the left side and $4$ on the right side of the equation to obtain: $7(2)+7(5x)=11+4(8x)+4(3) \\14+35x=11+32x+12 \\35x+14=32x+23$ Subtract $32x$ and $14$ to both sides of the equation: $35x+14-32x-14=32x+23-32x-14 \\3x=9$ Divide 3 to both sides: $\frac{3x}{3}=\frac{9}{3} \\x=3$
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