Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 6 - Section 6.6 - Solving Quadratic Equations by Factoring - Exercise Set: 53

Answer

The solutions are - $\frac{4}{3}$ and 5.

Work Step by Step

6$y^{2}$-22y-40=0 6$y^{2}$+8y-30y-40=0 2y(3y+4)-10(3y+4)=0 (2y-10)(3y+4)=0 2y-10 =0 or 3y+4=0 2y=10 0r y=-$\frac{4}{3}$ y=5 or y=-$\frac{4}{3}$ The solutions are - $\frac{4}{3}$ and 5. Check Let y=- $\frac{4}{3}$ 6$y^{2}$-22y-40=0 6$(-4/3)^{2}$-22$\frac{-4}{3}$-40=0 6*16/9+$\frac{88}{3}$-40=0 2*16/3+$\frac{88}{3}$-40=0 32/3-$+frac{88}{3}$-40=0 $\frac{120}{3}$-40=0 40-40=0 0=0 Let y=5 6$y^{2}$-22y-40=0 6$(5)^{2}$-22*5-40=0 6*25-110-40=0 150-110-40=0 0=0
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