College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding - Page 118: 74

Answer

$y=\frac{1}{2}, y=\frac{7}{6}$

Work Step by Step

$(\frac{y}{y-1})^{2}=6(\frac{y}{y-1})+7$ Simplify $\frac{y^{2}}{(y-1)^{2}}=\frac{6}{y-1}+7$ Put everything on the left side $\frac{y^{2}}{(y-1)^{2}}-\frac{6}{y-1}-7=0$ Add them $\frac{y^{2}-6y(y-1)-7(y-1)(y-1)}{(y-1)^{2}}=0$ Multiply both side by $(y-1)^{2}$ ${y^{2}-6y(y-1)-7(y-1)(y-1)}=0$ Simplify $12y^{2}-20y+7=0$ Use quadratic equation $\Delta= 20^{2}-7*12*4=64$ $\frac{20+\sqrt 64}{24}=\frac{7}{6}$ $\frac{20-\sqrt 64}{24}=\frac{1}{2}$
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