Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 57: 90

Answer

$x=-4$

Work Step by Step

$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{x^{2}-4}$ Factor the denominator of the second fraction of the right side of the equation: $\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{(x-2)(x+2)}$ Multiply the whole equation by $(x-2)(x+2)$: $(x+2)(x+5)=5(x-2)+28$ $x^{2}+7x+10=5x-10+28$ $x^{2}+7x+10=5x+18$ Take all terms to the left side: $x^{2}+7x+10-5x-18=0$ Simplify: $x^{2}+2x-8=0$ Solve by factoring: $(x+4)(x-2)=0$ Our two solutions are: $x=-4$ and $x=2$ If we check our answers, we'll see that the equation is undefined when we substitute $x=2$, so, our only solution is $x=-4$
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