Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.6 Other Types of Equations and Applications - 1.6 Exercises - Page 146: 14


The solution to the given equation cannot be $-3$ or $1$.

Work Step by Step

Factor the trinomial denominator to obtain: $\dfrac{2}{x+3}-\dfrac{5}{x-1}=\dfrac{-5}{(x+3)(x-1)}$ RECALL: The denominator of a rational expression is not allowed to be equal to zero. To find the numbers that will make the denominator equal to zero, equate each denominator to zero then solve each equation to obtain: \begin{array}{ccccccc} &x+3=0 &\text{or} &x-1=0 &\text{or} &(x+3)(x-1)=0 && \\&x=-3 &\text{or} &x=1 &\text{or} &x+3=0 &\text{or} &x-1=0 \\&x=-3 &\text{or} &x=1 &\text{or} &x=-3 &\text{or} &x=1 \end{array} Thus, the numbers $\left\{-3, 1\right\}$ are not possible solutions to the given equation. Therefore, the solution to the given equation cannot be $-3$ or $1$.
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