Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 4 - Section 4.5 - Linear Programming - Review Exercises - Page 309: 27

Answer

The solution set is $\{-\frac{11}{2},\frac{23}{2}\}$.

Work Step by Step

The given equation is $2\left | x-3 \right |-7=10$ Add $7$ to both sides. $2\left | x-3 \right |-7+7=10+7$ Simplify. $2\left | x-3 \right |=17$ Divide both sides by $2$. $\frac{2\left | x-3 \right |}{2}=\frac{17}{2}$ Simplify. $\left | x-3 \right |=\frac{17}{2}$ Remove the absolute value bars. $ x-3 =-\frac{17}{2}$ or $ x-3 =\frac{17}{2}$ Add $3$ to both sides. $ x-3+3 =-\frac{17}{2}+3$ or $ x-3+3 =\frac{17}{2}+3$ Simplify. $ x=\frac{-17+6}{2}$ or $ x =\frac{17+6}{2}$ $ x=\frac{-11}{2}$ or $ x =\frac{23}{2}$.
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