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: 68

Answer

$-12$

Work Step by Step

Multiply 12 (the LCD of the fractions) on both sides of the equation to obtain: $\require{cancel} 12(\frac{x}{4}) = 12(2+\frac{x-3}{3}) \\\frac{12x}{4}= 24+ 12(\frac{x-3}{3}) \\3x= 24+ \cancel{12}4(\frac{x-3}{\cancel{3}}) \\3x=24+4(x-3) \\3x=24+4x-12 \\3x=12+4x$ Subtract $3x$ on both sides of the equation to obtain: $0=12+x$ Subtract $12$ on both sides of the equation to obtain: $-12=x$
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