Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 2 - Solving Equations - 2-7 Solving Proportions - Practice and Problem-Solving Exercises - Page 129: 55



Work Step by Step

$\frac{4y-3}{y^2+1}=\frac{4}{y}$ cross multiply $y(4y-3)=4(y^2+1)$ distributive property $4y^2-3y=4y^2+4$ subtract $4y^2$ from each side $4y^2-3y-4y^2=4y^2+4-4y^2$ combine like terms $-3y=4$ divide both sides by -3 $-3y\div-3=4\div-3$ simplify $y=-1\frac{1}{3}$
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