Intermediate Algebra for College Students (7th Edition)

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

Chapter 6 - Section 6.6 - Rational Equations - Exercise Set - Page 465: 88

Answer

$\{15\}$.

Work Step by Step

Multiply both sides of the given equation by the Least Common Denominator $20$ of all denominators to clear fractions. $\Rightarrow 20\left( \frac{x-4}{2}-\frac{1}{5} \right)= 20\left( \frac{7x+1}{20} \right)$ Use the distributive property. $\Rightarrow 20\left( \frac{x-4}{2}\right)-20\left(\frac{1}{5} \right)= 20\left( \frac{7x+1}{20} \right)$ Cancel common factors. $\Rightarrow 10(x-4)-4(1)= 1\left( 7x+1\right)$ Use the distributive property. $\Rightarrow 10x-40-4= 7x+1$ Combine like terms. $\Rightarrow 10x-44= 7x+1$ Add $-7x+44$ to both sides. $\Rightarrow 10x-44-7x+44= 7x+1-7x+44$ Simplify. $\Rightarrow 3x= 45$ Divide both sides by $3$. $\Rightarrow \frac{3x}{3}= \frac{45}{3}$ Simplify. $\Rightarrow x= 15$ The solution set is $\{15\}$.
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