Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 4 - Section 4.8 - Solving Equations Containing Fractions - Exercise Set - Page 312: 67


The solution is 4

Work Step by Step

$\frac{b}{4}$ =$\frac{b}{12}$ + $\frac{2}{3}$ Multiply by LCD, 12 12$\frac{b}{4}$ =12$\frac{b}{12}$ + 12$\frac{2}{3}$ $\frac{4\times3 \times b}{4}$ =$\frac{12 \times b}{12}$ + $\frac{4\times3\times2}{3}$ 3b =b+8 3b-b=b+8-b 2b=8 $\frac{2b}{2}$=$\frac{8}{2}$ b=4 The solution is 4 Check Replace b with 4 $\frac{b}{4}$ =$\frac{b}{12}$ + $\frac{2}{3}$ $\frac{4}{4}$ =$\frac{4}{12}$ + $\frac{2}{3}$ LCD is 12 12$\frac{4}{4}$ =12$\frac{4}{12}$ +12 $\frac{2}{3}$ 12=4+ 8 12=12
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