College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.2 - Page 119: 53



Work Step by Step

We may substitute the expressions involving $x$ into the given equation $y_{1}-y_{2}=1$ and solve: $(\frac{x-3}{5})-(\frac{x-5}{4})=1$ $(\frac{4}{4})(\frac{x-3}{5})-(\frac{5}{5})(\frac{x-5}{4})=1$ (We multiply each of the fractions by a number equal to 1 so that the denominator of each becomes the least common denominator.) $(\frac{4x-12}{20})-(\frac{5x-25}{20})=1$ $\frac{(4x-12)-(5x-25)}{20}=1$ $\frac{4x-12-5x+25}{20}=1$ (Don't forget to distribute the negative.) $\frac{-x+13}{20}=1$ $(20)(\frac{-x+13}{20})=(20)(1)$ (Multiply both sides by 20 to eliminate the denominator.) $-x+13=20$ $-x=20-13$ $-x=7$ $x=-7$ Because we are dealing with a linear equality in one variable, there is only one answer.
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