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 - Study Summary - Practice Exercises - Page 146: 4


$t =-\dfrac{1}{10}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \dfrac{1}{6}t-\dfrac{3}{4}=t-\dfrac{2}{3} ,$ remove first the fraction by multiplying both sides by the $LCD.$ Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ The $LCD$ of the denominators, $\{ 6,4,1,3 \},$ is $ 12 $ since this is the least number that can be evenly divided (no remainder) by all the denominators. Multiplying both sides by the $LCD,$ the given equation is equivalent to \begin{array}{l}\require{cancel} 12\left( \dfrac{1}{6}t-\dfrac{3}{4} \right)=12\left( t-\dfrac{2}{3} \right) \\\\ 2t-9 =12t-8 .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} 2t-9 =12t-8 \\\\ 2t-12t =-8+9 \\\\ -10t =1 \\\\ t =\dfrac{1}{-10} \\\\ t =-\dfrac{1}{10} .\end{array}
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