Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 2 - Equations, Inequalities, and Problem Solving - 2.2 Using the Principles Together - 2.2 Exercise Set: 49



Work Step by Step

Multiplying both sides by the $LCD= 12 $, then the solution to the given equation, $ \dfrac{2}{3}+\dfrac{1}{4}t=2 $, is \begin{array}{l} 4(2)+3(t)=12(2) \\\\ 8+3t=24 \\\\ 3t=24-8 \\\\ 3t=16 \\\\ t=\dfrac{16}{3} .\end{array} CHECKING: \begin{array}{l} \dfrac{2}{3}+\dfrac{1}{4}\cdot\dfrac{16}{3}=2 \\\\ \dfrac{2}{3}+\dfrac{4}{3}=2 \\\\ \dfrac{6}{3}=2 \\\\ 2=2 \text{ (TRUE)} .\end{array} Hence, the solution is $ t=\dfrac{16}{3} $.
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