## College Algebra (11th Edition)

$x=\left\{ \dfrac{11}{27},\dfrac{25}{27} \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\left| \dfrac{7}{2-3x} \right|-9=0 ,$ use the properties of equality to isolate the absolute value expression. Then use the definition of absolute value equality. $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} \left| \dfrac{7}{2-3x} \right|=9 .\end{array} 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} \dfrac{7}{2-3x}=9 \\\\\text{OR}\\\\ \dfrac{7}{2-3x}=-9 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} \dfrac{7}{2-3x}=9 \\\\ (2-3x)\left( \dfrac{7}{2-3x}\right)=(9)(2-3x) \\\\ 1(7)=(9)(2-3x) \\\\ 7=18-27x \\\\ 27x=18-7 \\\\ 27x=11 \\\\ x=\dfrac{11}{27} \\\\\text{OR}\\\\ \dfrac{7}{2-3x}=-9 \\\\ (2-3x)\left( \dfrac{7}{2-3x} \right)=(-9)(2-3x) \\\\ 1(7)=(-9)(2-3x) \\\\ 7=-18+27x \\\\ -27x=-18-7 \\\\ -27x=-25 \\\\ x=\dfrac{-25}{-27} \\\\ x=\dfrac{25}{27} .\end{array} Hence, the solutions are $x=\left\{ \dfrac{11}{27},\dfrac{25}{27} \right\} .$