Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.1 Linear Equations - 1.1 Exercises - Page 92: 21



Work Step by Step

Multiply the LCD, which is $70$, to both sides of the equation to obtain: $\require{cancel}70 \cdot \frac{1}{14}(3x-2)=70 \cdot \frac{x+10}{10} \\5\cancel{70} \cdot \frac{1}{\cancel{14}}(3x-2)=7\cancel{70} \cdot \frac{x+10}{\cancel{10}} \\5(3x-2)=7(x+10)$ Distribute $5$ and $7$ to obtain: $5(3x)-5(2)=7(x) + 7(10) \\15x-10=7x+70$ Subtract $7x$ and add $10$ on both sides of the equation, then combine like terms to obtain: $15x-10-7x+10=7x+70-7x+10 \\8x=80$ Divide $8$ to both sides: $\dfrac{8x}{8} = \dfrac{80}{8} \\x=10$
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