Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.4 - Solving Linear Equations - Exercise Set - Page 50: 33



Work Step by Step

For easier work, it is better to get rid of the fractions. This can be achieved by multiplying the LCD of $12$ on both sides of the equation to obtain: $12(\frac{x+3}{6}) = 12(\frac{2}{3}+\frac{x-5}{4}) \\\frac{12(x+3)}{6} = \frac{24}{3} +\frac{12(x-5)}{4} \\2(x+3) = 8+3(x-5) \\2(x) + 2(3) = 8+3(x)-3(5) \\2x+6 = 8+3x-15 \\2x+6 = 3x-7$ Subtract $2x$ and add $7$ on both sides to obtain: $6+7 = 3x-2x \\13=x$ Check: $\begin{array}{ccc} &\frac{13+3}{6} &= &\frac{2}{3}+\frac{13-5}{4} \\&\frac{16}{6} &= &\frac{2}{3}+\frac{8}{4} \\&\frac{8}{3} &= &\frac{2}{3} + 2 \\&\frac{8}{3} &= &\frac{2}{3} + \frac{6}{3} \\&\frac{8}{3} &= &\frac{8}{3}\end{array}$
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