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

$x=11$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\dfrac{1}{3}(2x-1)=7 ,$ use the properties of equality to isolate the variable. Do checking of the solution. $\bf{\text{Solution Details:}}$ Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{3}(2x-1)=7 \\\\ 3\cdot\dfrac{1}{3}(2x-1)=3\cdot7 \\\\ 2x-1=21 \\\\ 2x=21+1 \\\\ 2x=22 \\\\ x=\dfrac{22}{2} \\\\ x=11 .\end{array} Checking: If $x=11,$ then \begin{array}{l}\require{cancel} \dfrac{1}{3}(2x-1)=7 \\\\ \dfrac{1}{3}(2\cdot11-1)=7 \\\\ \dfrac{1}{3}(22-1)=7 \\\\ \dfrac{1}{3}(21)=7 \\\\ 7=7 \text{ (TRUE) } .\end{array} Hence, the solution is $x=11 .$