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 697: 46

Answer

$9$.

Work Step by Step

The given expression is $\Rightarrow \frac{x-6}{x+3}+\frac{2x}{x-3}=\frac{4x+3}{x+3}$ Multiply the equation by $(x+3)(x-3)$. $\Rightarrow (x+3)(x-3)\left ( \frac{x-6}{x+3}+\frac{2x}{x-3} \right )=(x+3)(x-3)\left (\frac{4x+3}{x+3}\right )$ Use the distributive property and divide out common factors. $\Rightarrow (x-3)(x-6)+2x( x+3 )=(x-3)(4x+3)$ Simplify. $\Rightarrow x^2-9x+18+2x^2+6x=4x^2-9x-9$ Move all terms to the one side. $\Rightarrow x^2-9x+18+2x^2+6x-4x^2+9x+9=0$ Add like terms. $\Rightarrow -x^2+6x+27=0$ Multiply by $-1$. $\Rightarrow x^2-6x-27=0$ Factor. $\Rightarrow (x-9)(x-3)=0$ Use zero product property. $x-9=0$ or $x-3=0$ Solve for $x$. $x=9$ or $x=3$ Check $x=9$. $\Rightarrow \frac{9-6}{9+3}+\frac{2(9)}{9-3}=\frac{4(9)+3}{9+3}$ $\Rightarrow \frac{3}{12}+\frac{18}{6}=\frac{39}{12}$ $\Rightarrow \frac{1}{4}+\frac{3}{1}=\frac{13}{4}$ $\Rightarrow \frac{1+12}{4}=\frac{13}{4}$ $\Rightarrow \frac{13}{4}=\frac{13}{4}$ Check $x=3$. $\Rightarrow \frac{3-6}{3+3}+\frac{2(3)}{3-3}=\frac{4(3)+3}{3+3}$ $\Rightarrow \frac{-3}{6}+\frac{6}{0}=\frac{15}{6}$ Undefined. Hence, the correct solution is $x=9$
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