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: 34



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: $12(\frac{x+1}{4}) = 12(\frac{1}{6}+\frac{2-x}{3}) \\\frac{12(x+1)}{4} = \frac{12}{6} +\frac{12(2-x)}{3} \\3(x+1) = 2+4(2-x) \\3(x)+3(1) = 2+4(2) - 4(x) \\3x+3=2+8-4x \\3x+3=10-4x$ Add $4x$ and subtract $3$ on both sides to obtain: $3x+4x=10-3 \\7x=7 \\\frac{7x}{7}=\frac{7}{7} \\x = 1$ Check: $\begin{array}{ccc} &\frac{1+1}{4} &= &\frac{1}{6}+\frac{2-1}{3} \\&\frac{2}{4} &= &\frac{1}{6}+\frac{1}{3} \\&\frac{1}{2} &= &\frac{1}{6} + \frac{2}{6} \\&\frac{1}{2} &= &\frac{3}{6} \&\frac{1}{2} &= &\frac{1}{2}\end{array}$
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