Intermediate Algebra for College Students (7th Edition)

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

Chapter 7 - Section 7.7 - Complex Numbers - Exercise Set - Page 572: 146

Answer

$\{\frac{21}{11}\}$.

Work Step by Step

The given equation is $\Rightarrow 2x-\frac{x-3}{8}=\frac{1}{2}+\frac{x+5}{2}$ The LCD is $8$. Multiply the equation by $8$. $\Rightarrow 8\cdot \left (2x-\frac{x-3}{8}\right )=8\cdot \left (\frac{1}{2}+\frac{x+5}{2}\right )$ Use the distributive property. $\Rightarrow 8\cdot 2x-8\cdot \frac{x-3}{8}=8\cdot \frac{1}{2}+8\cdot \frac{x+5}{2}$ Simplify. $\Rightarrow 16x- (x-3)=4+4\cdot (x+5)$ Use the distributive property. $\Rightarrow 16x- x+3=4+4x+20$ Add like terms. $\Rightarrow 15x+3=24+4x$ Add $-4x-3$ to both sides. $\Rightarrow 15x+3-4x-3=24+4x-4x-3$ Simplify. $\Rightarrow 11x=21$ Divide both sides by $11$. $\Rightarrow \frac{11x}{11}=\frac{21}{11}$ Simplify. $\Rightarrow x=\frac{21}{11}$ The solution set is $\{\frac{21}{11}\}$.
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