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: 35



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: $12(\frac{x}{4}) = 12(2+\frac{x-3}{3}) \\\frac{12(x)}{4} = 12(2) +\frac{12(x-3)}{3} \\3x = 24 + 4(x-3) \\3x = 24+4x-12 \\3x=4x+12$ Subtract $3x$ and $12$ on both sides to obtain: $-12=4x-3x \\-12=x$ Check: $\begin{array}{ccc} &\frac{-12}{4} &= &2+\frac{-12-3}{3} \\&-3 &= &2+\frac{-15}{3} \\&-3 &= &2 + (-5) \\&-3 &= &-3\end{array}$
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