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 - Mid-Chapter Check Point - Page 53: 4

Answer

x = -7

Work Step by Step

$\frac{x-3}{5}$ - $1$= $\frac{x-5}{4}$ Find the least common multiple: 5 : $5$ 4 : $2\times2$ Hence, the LCM is $5\times2\times2 = 20$ Multiply the whole equation by the LCM: $[\frac{x-3}{5} - 1]\times20$ = $[\frac{x-5}{4}]\times20$ $4(x-3) - 20$ = $5(x-5)$ Expand $4(x-3)$ = $4x - 12$ Expand $5(x-5)$ = $5x-25$ Rewrite the equation: $4x - 12 -20$ = $5x-25$ $4x -32$ = $5x-25$ Add 32 to both sides: $4x -32 + 32$ = $5x-25 + 32$ $4x$ = $5x+7$ Simplify: $x=-7$
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