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: 9

Answer

$x=6$

Work Step by Step

$\sqrt (2x-3)-2=1$ Add 2 to both sides of the equation. $\sqrt (2x-3)=3$ Square both sides. $(\sqrt (2x-3))^{2}=3^{2}$ $2x-3=9$ Add 3 to both sides. $2x=12$ Divide both sides by 2. $x=6$
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