College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.6 - Page 179: 94


$x=-9$ and $x=\dfrac{3}{4}$ satisfy the given conditions

Work Step by Step

$y_{1}=6\Big(\dfrac{2x}{x-3}\Big)^{2}$, $y_{2}=5\Big(\dfrac{2x}{x-3}\Big)$, and $y_{1}$ exceeds $y_{2}$ by $6$ If $y_{1}$ exceeds $y_{2}$ by $6$, then $y_{2}+6=y_{1}$ and $y_{1}-y_{2}=6$ If $y_{1}-y_{2}=6$, then $6\Big(\dfrac{2x}{x-3}\Big)^{2}-5\Big(\dfrac{2x}{x-3}\Big)$ is also equal to $6$: $6\Big(\dfrac{2x}{x-3}\Big)^{2}-5\Big(\dfrac{2x}{x-3}\Big)=6$ Evaluate all operations indicated on the left side: $6\Big[\dfrac{4x^{2}}{(x-3)^{2}}\Big]-\dfrac{10x}{x-3}=6$ $\dfrac{24x^{2}}{(x-3)^{2}}-\dfrac{10x}{x-3}=6$ Multiply the whole equation by $(x-3)^{2}$ $(x-3)^{2}\Big[\dfrac{24x^{2}}{(x-3)^{2}}-\dfrac{10x}{x-3}=6\Big]$ $24x^{2}-10x(x-3)=6(x-3)^{2}$ Evaluate all indicated operations: $24x^{2}-10x^{2}+30x=6(x^{2}-6x+9)$ $24x^{2}-10x^{2}+30x=6x^{2}-36x+54$ Take all terms to the left side of the equation and simplify: $24x^{2}-10x^{2}+30x-6x^{2}+36x-54=0$ $8x^{2}+66x-54=0$ Solve by factoring: $2(x+9)(4x-3)=0$ Take the $2$ to divide the right side: $(x+9)(4x-3)=\dfrac{0}{2}$ $(x+9)(4x-3)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x+9=0$ $x=-9$ $4x-3=0$ $4x=3$ $x=\dfrac{3}{4}$ $x=-9$ and $x=\dfrac{3}{4}$ satisfy the given conditions
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