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: 21

Answer

{$\frac{1}{2},4$}

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 $x(x-3)$: $\frac{5}{x-3}+\frac{4}{x}=6$ $\frac{5(x)+4(x-3)}{x(x-3)}=6$ $\frac{5x+4x-12}{x(x-3)}=6$ $\frac{9x-12}{(x^{2}-3x)}=6$ Now, we cross multiply the two fractions in order to create a quadratic equation: $\frac{9x-12}{(x^{2}-3x)}=\frac{6}{1}$ $1(9x-12)=6(x^{2}-3x)$ $9x-12=6x^{2}-18x$ $6x^{2}-18x=9x-12$ $6x^{2}-18x-9x+12=0$ $6x^{2}-27x+12=0$ $2x^{2}-9x+4=0$ Now, we use rules of factoring trinomials to solve the equation: $2x^{2}-9x+4=0$ $2x^{2}-8x-1x+4=0$ $2x(x-4)-1(x-4)=0$ $(x-4)(2x-1)=0$ $(x-4)=0$ or $(2x-1)=0$ $x=4$ or $x=\frac{1}{2}$ Therefore, the solution is {$\frac{1}{2},4$}.
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