Prealgebra (7th Edition)

Published by Pearson
ISBN 10: 0321955048
ISBN 13: 978-0-32195-504-3

Chapter 3 - Section 3.3 - Solving Linear Equations in One Variable - Exercise Set: 40

Answer

The solution is -10

Work Step by Step

5( 3y-2) = 16y Apply the distributive property 5$\times$3y - 5$\times$2 = 16y Multiply 15y -10 =16y Subtract 16y from both sides 15y -10 -16y =16y -16y -y -10= 0 -y -10+10= 0 +10 Add 10 to both sides -y = 10 Divide by -1 on both sides $\frac{-y}{-1}$ = $\frac{10}{-1}$ x = -10 The solution is -10 Check 5( 3y-2) = 16y 5[3(-10)-2] = 16(-10) 5[-30 -2]=-160 5[-32]=-160 -160 =-160
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