Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 10 - Quadratic Equations - Chapter 10 Review Problem Set - Page 468: 22



Work Step by Step

First, we add the fractions on the left hand side by taking their LCM. Upon inspection, the LCM is found to be $x(x+2)$: $\frac{1}{x+2}-\frac{2}{x}=3$ $\frac{1(x)-2(x+2)}{x(x+2)}=3$ $\frac{x-2x-4}{x(x+2)}=3$ $\frac{-x-4}{(x^{2}+2x)}=3$ Now, we cross multiply the two fractions in order to create a linear equation: $\frac{-x-4}{(x^{2}+2x)}=\frac{3}{1}$ $1(-x-4)=3(x^{2}+2x)$ $-x-4=3x^{2}+6x$ $3x^{2}+6x=-x-4$ $3x^{2}+6x+x+4=0$ $3x^{2}+7x+4=0$ Now, we use rules of factoring trinomials to solve the equation: $3x^{2}+7x+4=0$ $3x^{2}+3x+4x+4=0$ $3x(x+1)+4(x+1)=0$ $(x+1)(3x+4)=0$ $(x+1)=0$ or $(3x+4)=0$ $x=-1$ or $x=-\frac{4}{3}$ Therefore, the solution is {$-\frac{4}{3},-1$}.
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