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 - Page 203: 33


Conditional equation. $x=\frac{4}{7}$

Work Step by Step

Given equation, $\frac{4}{x+2} + \frac{3}{x} = \frac{10}{x^{2}+2x} $ $x=0 $ or $ x=-2$ makes the denominator zero, so, $\frac{4}{x+2} + \frac{3}{x} = \frac{10}{x^{2}+2x} ; x\ne0,-2$ Taking LCD, $\frac{4x+3(x+2)}{x(x+2)} = \frac{10}{x^{2}+2x} ; x\ne0,-2$ $\frac{4x+3x+6}{x(x+2)} = \frac{10}{x^{2}+2x} ; x\ne0,-2$ $\frac{7x+6}{x^{2}+2x} = \frac{10}{x^{2}+2x} ; x\ne0,-2$ Multiply both sides by $x^{2}+2x$ $(x^{2}+2x)\frac{7x+6}{x^{2}+2x} = (x^{2}+2x)\frac{10}{x^{2}+2x} ; x\ne0,-2$ $7x+6=10 ; x\ne0,-2$ $7x=10-6 ; x\ne0,-2$ $7x=4; x\ne0,-2$ $x=\frac{4}{7}; x\ne0,-2$ This equation has one solution, so it is a conditional equation.
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