Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.4 - Solving Linear Equations - Exercise Set - Page 53: 105

Answer

$12$

Work Step by Step

Since, we have as per the statement $3a-4=32$ Here, $a$ is any number. or,$3a=32+4$ or, $3a=36$ or, $a=\dfrac{36}{3} \implies a=12$ Hence, the result is $12$
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