#### Answer

$x=1$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\dfrac{1}{3}x+\dfrac{2}{5}=\dfrac{4}{5}+\dfrac{3}{5}x-\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, $\{
3,5
\}$ is $
15
$ since it is the least number that can be divided evenly (no remainder) by all the denominators.
Multiplying both sides by the $LCD=
15
$ and using the properties of equality results to
\begin{array}{l}\require{cancel}
\dfrac{1}{3}x+\dfrac{2}{5}=\dfrac{4}{5}+\dfrac{3}{5}x-\dfrac{2}{3}
\\\\
15\left( \dfrac{1}{3}x+\dfrac{2}{5} \right) =15\left( \dfrac{4}{5}+\dfrac{3}{5}x-\dfrac{2}{3} \right)
\\\\
5x+6=12+9x-10
\\\\
5x-9x=12-10-6
\\\\
-4x=-4
\\\\
x=\dfrac{-4}{-4}
\\\\
x=1
.\end{array}
Checking: If $x=1,$ then
\begin{array}{l}\require{cancel}
\dfrac{1}{3}x+\dfrac{2}{5}=\dfrac{4}{5}+\dfrac{3}{5}x-\dfrac{2}{3}
\\\\
\dfrac{1}{3}(1)+\dfrac{2}{5}=\dfrac{4}{5}+\dfrac{3}{5}(1)-\dfrac{2}{3}
\\\\
\dfrac{1}{3}+\dfrac{2}{5}=\dfrac{4}{5}+\dfrac{3}{5}-\dfrac{2}{3}
\\\\
\dfrac{5}{15}+\dfrac{6}{15}=\dfrac{12}{15}+\dfrac{9}{15}-\dfrac{10}{15}
\\\\
\dfrac{11}{15}=\dfrac{11}{15}
\text{ (TRUE) }
.\end{array}
Hence, the solution is $
x=1
.$