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

$t=\left\{ -\dfrac{1}{7},\dfrac{7}{3} \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $|5t-3|=2t+4 ,$ use the definition of an absolute value equality. $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|=c$ implies $x=c \text{ or } x=-c,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 5t-3=2t+4 \\\\\text{OR}\\\\ 5t-3=-(2t+4) .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 5t-3=2t+4 \\\\ 5t-2t=4+3 \\\\ 3t=7 \\\\ t=\dfrac{7}{3} \\\\\text{OR}\\\\ 5t-3=-(2t+4) \\\\ 5t-3=-2t-4 \\\\ 5t+2t=-4+3 \\\\ 7t=-1 \\\\ t=-\dfrac{1}{7} .\end{array} Hence, $t=\left\{ -\dfrac{1}{7},\dfrac{7}{3} \right\} .$