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

$t=-\dfrac{1}{5}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $|6-5t|=|5t+8| ,$ use the definition of an absolute value equality. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Since $|x|=|y|$ implies $x=y \text{ or } x=-y,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 6-5t=5t+8 \\\\\text{OR}\\\\ 6-5t=-(5t+8) .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 6-5t=5t+8 \\\\ -5t-5t=8-6 \\\\ -10t=2 \\\\ t=\dfrac{2}{-10} \\\\ t=-\dfrac{1}{5} \\\\\text{OR}\\\\ 6-5t=-(5t+8) \\\\ 6-5t=-5t-8 \\\\ -5t+5t=-8-6 \\\\ 0=-14 \text{ (FALSE)} .\end{array} Hence, $t=-\dfrac{1}{5} .$