Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.8 - Solving Equations by Factoring and Problem Solving - Exercise Set - Page 323: 70


3/2, 5/2

Work Step by Step

x+y=4 (x)(y)=15/4 x=4-y Substitute (4-y)(y)=15/4 4y-y^2=15/4 Rearrange 4y-y^2-15/4=0 or y^2-4y+15/4=0 In order to get rid of denominator 4, multiply the variables and constant by 4 4(y^2-4y+15/4)=0 4y^2-16y+15=0 Using ac method a=60 b=-16 The two numbers are -10 and -6 then 4y^2-10y-6y+15=0 2y(2y-5)-3(2y-5)=0 (2y-3)(2y-5)=0 2y-3=0 2y=3 y=3/2 2y-5=0 2y=5 y=5/2 3/2+5/2=4 (3/2)(5/2)=15/4
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