Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 11 - Rational Expressions and Functions - 11-5 Solving Rational Equations - Practice and Problem-Solving Exercises - Page 695: 19


The equation has one solution, $\frac{16}{3}$.

Work Step by Step

$\frac{4}{3(c+4)}+1=\frac{2c}{c+4}$ $\frac{4}{3(c+4)}+\frac{3(c+4)}{3(c+4)}-\frac{3\times2c}{3(c+4)}=0$ $\frac{4}{3(c+4)}+\frac{3c+12}{3(c+4)}-\frac{6c}{3(c+4)}=0$ $\frac{4+3c+12-6c}{3(c+4)}=0$ $16-3c=0\times3(c+4)$ $16-3c=0$ $3c=16$ $c=\frac{16}{3}$ Check: $4\div[3(16/3+4)]+1=4/28+1=8/7;(2\times16/3)\div(16/3+4)=(32/3)\div(28/3)=32/28=8/7$ The equation has one solution, $\frac{16}{3}$.
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