Answer
$x=-\dfrac{1}{3}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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}
.$
