Elementary Algebra

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

Chapter 10 - Quadratic Equations - Chapter 10 Test - Page 469: 14



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 $x(x-1)$: $\frac{2}{x-1}+\frac{1}{x}=\frac{5}{2}$ $\frac{2(x)+1(x-1)}{x(x-1)}=\frac{5}{2}$ $\frac{2x+x-1}{x(x-1)}=\frac{5}{2}$ $\frac{3x-1}{(x^{2}-x)}=\frac{5}{2}$ Now, we cross multiply the two fractions in order to create a linear equation: $\frac{3x-1}{x(x-1)}=\frac{5}{2}$ $2(3x-1)=5(x^{2}-x)$ $6x-2=5x^{2}-5x$ $5x^{2}-5x=6x-2$ $5x^{2}-5x-6x+2=0$ $5x^{2}-11x+2=0$ Now, we use rules of factoring trinomials to solve the equation: $5x^{2}-11x+2=0$ $5x^{2}-10x-1x+2=0$ $5x(x-2)-1(x-2)=0$ $(x-2)(5x-1)=0$ $(x-2)=0$ or $(5x-1)=0$ $x=2$ or $x=\frac{1}{5}$ Therefore, the solution is {$\frac{1}{5},2$}.
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