College Algebra (6th Edition)

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

Chapter 1 - Summary, Review, and Test - Review Exercises: 35

Answer

Conditional equation. $x=2$

Work Step by Step

Given equation, $\frac{x+2}{x+3} + \frac{1}{x^{2}+2x-3} - 1 = 0$ By factoring, $x^{2}+2x-3=(x+3)(x-1)$, the given equation can be written as $\frac{x+2}{x+3} + \frac{1}{(x+3)(x-1)} - 1 = 0$ $\frac{x+2}{x+3} + \frac{1}{(x+3)(x-1)} = 1$ $x=1 $ or $ x=-3$ makes the denominator zero, so, $\frac{x+2}{x+3} + \frac{1}{(x+3)(x-1)} = 1 ; x\ne 1,-3;$ Taking LCD, $ \frac{(x+2)(x-1)+1}{(x+3)(x-1)} = 1 ; x\ne 1,-3;$ $ \frac{(x^{2}+2x-x-2)+1}{(x+3)(x-1)} = 1 ; x\ne 1,-3;$ $ \frac{x^{2}+x-1}{(x+3)(x-1)} = 1 ; x\ne 1,-3;$ $ x^{2}+x-1 = (x+3)(x-1) ; x\ne 1,-3;$ $ x^{2}+x-1 = x^{2}+2x-3 ; x\ne 1,-3;$ $ x^{2}+x-1 -x^{2}-2x+3=0 ; x\ne 1,-3;$ $-x+2=0 ; x\ne 1,-3;$ $x=2 ; x\ne 1,-3;$ This equation has one solution, so it is a conditional equation.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.