Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.6 - Radical Equations and Problem Solving - Exercise Set: 10

Answer

$x=47$

Work Step by Step

$\sqrt (3x+3)-4=8$ Add 4 to both sides of the equation. $\sqrt (3x+3)=12$ Square both sides. $(\sqrt (3x+3))^{2}=12^{2}$ $3x+3=144$ Subtract 3 from both sides. $3x=141$ Divide both sides by 3. $x=47$
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