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

$x=-\dfrac{1}{3}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $-\dfrac{5}{6}+x=-\dfrac{1}{2}-\dfrac{2}{3} ,$ use the properties of equality to isolate the variable. Then do checking of the solution. $\bf{\text{Solution Details:}}$ The $LCD$ of the denominators, $\{ 6,2,3 \}$ is $6$ since it is the least number that can be divided evenly (no remainder) by all the denominators. Multiplying both sides by the $LCD= 6$ and using the properties of equality results to \begin{array}{l}\require{cancel} -\dfrac{5}{6}+x=-\dfrac{1}{2}-\dfrac{2}{3} \\\\ 6\left( -\dfrac{5}{6}+x \right)=6\left( -\dfrac{1}{2}-\dfrac{2}{3} \right) \\\\ -5+6x=-3-4 \\\\ 6x=-3-4+5 \\\\ 6x=-2 \\\\ x=-\dfrac{2}{6} \\\\ x=-\dfrac{1}{3} .\end{array} Checking: If $x=-\dfrac{1}{3},$ then \begin{array}{l}\require{cancel} -\dfrac{5}{6}+x=-\dfrac{1}{2}-\dfrac{2}{3} \\\\ -\dfrac{5}{6}-\dfrac{1}{3}=-\dfrac{1}{2}-\dfrac{2}{3} \\\\ -\dfrac{5}{6}-\dfrac{2}{6}=-\dfrac{3}{6}-\dfrac{4}{6} \\\\ -\dfrac{7}{6}=-\dfrac{7}{6} \text{ (TRUE) } .\end{array} Hence, the solution is $x=-\dfrac{1}{3} .$