Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 1 - Section 1.3 - Applications of Linear Equations - 1.3 Exercises: 25

Answer

$\dfrac{1}{3}x-\dfrac{13}{2}$

Work Step by Step

Without an equal sign, the given is an $\text{ expression }.$ By combining like terms, the given expression, $ \dfrac{1}{2}x-\dfrac{1}{6}x+\dfrac{3}{2}-8 ,$ simplifies to \begin{array}{l}\require{cancel} \left( \dfrac{1}{2}x-\dfrac{1}{6}x \right)+\left( \dfrac{3}{2}-8 \right) \\\\= \left( \dfrac{3}{6}x-\dfrac{1}{6}x \right)+\left( \dfrac{3}{2}-\dfrac{16}{2} \right) \\\\= \left( \dfrac{3-1}{6}x\right)+\left( \dfrac{3-16}{2}\right) \\\\= \left( \dfrac{2}{6}x\right)+\left( \dfrac{-13}{2}\right) \\\\= \dfrac{1}{3}x-\dfrac{13}{2} .\end{array}
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