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: 55

Answer

$x=2$

Work Step by Step

We plug the expressions involving $x$ into the given equation $y_{1}+y_{2}=y_{3}$, look for a common denominator, and solve: $(\frac{5}{x+4})+(\frac{3}{x+3})=\frac{12x+19}{x^{2}+7x+12}$ $(\frac{5}{x+4})+(\frac{3}{x+3})=\frac{12x+19}{(x+3)(x+4)}$ (Notice that the denominator on the right-hand side of the equation factors nicely.) $(\frac{x+3}{x+3})(\frac{5}{x+4})+(\frac{x+4}{x+4})(\frac{3}{x+3})=\frac{12x+19}{(x+3)(x+4)}$ $\frac{5(x+3)}{(x+3)(x+4)}+\frac{3(x+4)}{(x+3)(x+4)}=\frac{12x+19}{(x+3)(x+4)}$ $\frac{5x+15+3x+12}{(x+3)(x+4)}=\frac{12x+19}{(x+3)(x+4)}$ $\frac{8x+27}{(x+3)(x+4)}=\frac{12x+19}{(x+3)(x+4)}$ $(x+3)(x+4)(\frac{8x+27}{(x+3)(x+4)})=(x+3)(x+4)(\frac{12x+19}{(x+3)(x+4)})$ $8x+27=12x+19$ $27-19=12x-8x$ $4x=8$ $x=2$
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