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: 59

Answer

The solution is 4

Work Step by Step

3(5c-1)-2=13c+3 Apply the distributive property 3$\times$ 5c - 3$\times$1 -2=13c+3 Multiply 15c-3-2=13c+3 Simplify 15c-5=13c+3 Add 5 to both sides . 15c-5 +5=13c+3 +5 Simplify 15c=13c+8 Subtract 13c from both sides 15c-13c=13c+8 -13c 2c=8 Divide by 2 on both sides c=4 The solution is 4 Check 3(5c-1)-2=13c+3 Replace c with 4 3(5$\times$4 -1)-2=13$\times$4 +3 3(20-1)-2=52+3 3$\times$19-2=55 57-2=55 55=55
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