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

$x=-\dfrac{5}{32}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\dfrac{2}{3}\left( \dfrac{7}{8}-4x \right)-\dfrac{5}{8}=\dfrac{3}{8} ,$ remove first the fraction by multiplying both sides by the $LCD.$ Then use the properties of equality to isolate the variable. Do checking of the solution. $\bf{\text{Solution Details:}}$ The $LCD$ of the denominators, $\{ 3,8,8 \},$ is $24$ since this is the least number that can be evenly divided (no remainder) by all the denominators. Multiplying both sides by the $LCD,$ the given equation is equivalent to \begin{array}{l}\require{cancel} \dfrac{2}{3}\left( \dfrac{7}{8}-4x \right)-\dfrac{5}{8}=\dfrac{3}{8} \\\\ 24\left[\dfrac{2}{3}\left( \dfrac{7}{8}-4x \right)-\dfrac{5}{8}\right]=24\left[\dfrac{3}{8}\right] \\\\ 16\left( \dfrac{7}{8}-4x \right)-15=9 .\end{array} Using the Distributive Property and the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} 16\left( \dfrac{7}{8}-4x \right)-15=9 \\\\ 14-64x-15=9 \\\\ -64x=9-14+15 \\\\ -64x=10 \\\\ x=\dfrac{10}{-64} \\\\ x=-\dfrac{5}{32} .\end{array} Checking: If $x=-\dfrac{5}{32},$ then \begin{array}{l}\require{cancel} \dfrac{2}{3}\left( \dfrac{7}{8}-4x \right)-\dfrac{5}{8}=\dfrac{3}{8} \\\\ \dfrac{2}{3}\left( \dfrac{7}{8}-4\left(-\dfrac{5}{32}\right) \right)-\dfrac{5}{8}=\dfrac{3}{8} \\\\ \dfrac{2}{3}\left( \dfrac{7}{8}+\dfrac{5}{8} \right)-\dfrac{5}{8}=\dfrac{3}{8} \\\\ \dfrac{2}{3}\left( \dfrac{12}{8}\right)-\dfrac{5}{8}=\dfrac{3}{8} \\\\ \dfrac{2}{\cancel3}\left( \dfrac{\cancel3(4)}{8}\right)-\dfrac{5}{8}=\dfrac{3}{8} \\\\ \dfrac{8}{8}-\dfrac{5}{8}=\dfrac{3}{8} \\\\ \dfrac{3}{8}=\dfrac{3}{8} \text{ (TRUE) } .\end{array} Hence, the solution is $x=-\dfrac{5}{32} .$