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: 29



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 $15$ on both sides of the equation to obtain: $15(\frac{3x}{5}) = 15(\frac{2x}{3}+1) \\\frac{45x}{5} - \frac{30x}{3} +15 \\9x = 10x+15$ Subtract $9x$ and $15$ on both sides to obtain: $-15=10x-9x \\-15=x$ Check: $\begin{array}{ccc} &\frac{3(-15)}{5} &= &\frac{2(-15)}{3}+1 \\&\frac{-45}{5} &= &\frac{-30}{3}+1 \\&-9 &= &-10+1 \\&-9 &= &-9\end{array}$
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