Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 11 - Rational Expressions and Functions - 11-5 Solving Rational Equations - Practice and Problem-Solving Exercises - Page 695: 31


The equation has two solutions, $\frac{-2}{3}$ and $3$.

Work Step by Step

To use the "cross products property" to solve an equation, that equation must be written in the form of two equivalent fractions (one fraction on each side). In this equation: $\frac{1}{x-2}=\frac{2x-6}{x+6}+1$, so we can see that the left side has only one fraction, which satisfies our requirement. But the right side contains more than a single fraction, which makes us unable to use the "cross products property". Therefore, we need to add the sum up so that the right side only contains one fraction. $\frac{2x-6}{x+6}+1=\frac{2x-6}{x+6}+\frac{x+6}{x+6}=\frac{2x-6+x+6}{x+6}=\frac{3x}{x+6}$ Rewriting the equation, we have: $\frac{1}{x-2}=\frac{3x}{x+6}$ Now we can use the "cross products property": $1(x+6)=3x(x-2)$ $x+6=3x^2-6x$ $3x^2-7x-6=0$ $3x^2-9x+2x-6=0$ $3x(x-3)+2(x-3)=0$ $(3x+2)(x-3)=0$ $3x+2=0$ or $x-3=0$ $x=-\frac{2}{3}$ or $x=3$ Check: $\frac{1}{-2/3-2}=\frac{-3}{8}; \frac{2\times-2/3-6}{-2/3+6}+1=-\frac{11}{8}+1=\frac{3}{8}$ $\frac{1}{3-2}=1; \frac{2\times3-6}{3+6}+1=-\frac{0}{9}+1=1$ The equation has two solutions, $\frac{-2}{3}$ and $3$.
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