## Precalculus (6th Edition)

$\color{blue}{\left\{\frac{11}{27}, \frac{25}{27}\right\}}$
Add $9$ to both sides of the equation to obtain: $\left|\dfrac{7}{2-3x}\right|-9+9=0+9 \\\left|\dfrac{7}{2-3x}\right|=9$ RECALL: $|x|=a \longrightarrow x=a \text{ or } x = -a$ Use the rule above to have: $\left|\dfrac{7}{2-3x}\right|= 9 \longrightarrow \dfrac{7}{2-3x}=9 \text{ or } \dfrac{7}{2-3x} = -9$ Solve each equation to obtain: $\dfrac{7}{2-3x}=9 \\7=9(2-3x) \\7=9(2)-9(3x) \\7=18-27x \\7-18=-27x \\-11=-27x \\-\frac{11}{-27}=-\frac{27x}{-27} \\\frac{11}{27}=x$ or $\dfrac{7}{2-3x}=-9 \\7=-9(2-3x) \\7=-9(2)-(-9)(3x) \\7=-18+27x \\7+18=27x \\25=27x \\\frac{25}{27}=\frac{27x}{27} \\\frac{25}{27}=x$ Thus, the solution set is $\color{blue}{\left\{\frac{11}{27}, \frac{25}{27}\right\}}$.