Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.4 Equations - R.4 Exercises: 32

Answer

The solution is $y=-\dfrac{5}{2}$

Work Step by Step

$\dfrac{2y}{y-1}=\dfrac{5}{y}+\dfrac{10-8y}{y^{2}-y}$ Take out common factor $y$ from the denominator of the second term on the right side of the equation: $\dfrac{2y}{y-1}=\dfrac{5}{y}+\dfrac{10-8y}{y(y-1)}$ Multiply the whole equation by $y(y-1)$ $y(y-1)\Big[\dfrac{2y}{y-1}=\dfrac{5}{y}+\dfrac{10-8y}{y(y-1)}\Big]$ $2y(y)=5(y-1)+10-8y$ $2y^{2}=5y-5+10-8y$ Take all terms to the left side of the equation and simplify: $2y^{2}-5y+8y+5-10=0$ $2y^{2}+3y-5=0$ Solve by factoring: $(2y+5)(y-1)=0$ Set both factors equal to $0$ and solve each individual equation for $y$: $2y+5=0$ $2y=-5$ $y=-\dfrac{5}{2}$ $y-1=0$ $y=1$ The original equation is undefined for $y=1$, so the solution is $y=-\dfrac{5}{2}$
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