Intermediate Algebra for College Students (7th Edition)

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

Chapter 1 - Review Exercises: 69

Answer

$\frac{77}{15}$

Work Step by Step

Multiply 12 (the LCD of the fractions) on both sides of the equation to obtain: $\require{cancel} 12(\frac{3x+1}{3}-\frac{13}{2}) = 12(\frac{1-x}{4}) \\12(\frac{3x+1}{3})-12(\frac{13}{2}) = \cancel{12}3(\frac{1-x}{\cancel{4}}) \\\cancel{12}4(\frac{3x+1}{\cancel{3}})-\cancel{12}6(\frac{13}{\cancel{2}}) = 3(1-x) \\4(3x+1)-6(13)=3-3x \\12x+4-78=3-3x \\12x-74=3-3x$ Add $3x$ on both sides of the equation to obtain: $15x-74=3$ Add $74$ on both sides of the equation to obtain: $15x=77$ Divide 15 on both sides of the equation to obtain: $x=\frac{77}{15}$
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