Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.7 - Equations - Concept and Vocabulary Check - Page 105: 7

Answer

$5(x+3)+3(x+4)=12x+9$

Work Step by Step

On the LHS, $(x+4)$ cancels when multiplying $\displaystyle \frac{5}{x+4}$ with $(x+4)(x+3)$ What is left is $5(x+3)$ $(x+3)$ cancels when multiplying $\displaystyle \frac{3}{x+3}$ with $(x+4)(x+3)$ What is left is $3(x+4)$ So, LHS = $5(x+3)+3(x+4)$ On the RHS, both parentheses cancel, and we have RHS = $12x+9$ The equation we work further with is $5(x+3)+3(x+4)=12x+9$
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