#### Answer

$y=-3$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
1-\dfrac{2}{3}y=\dfrac{9}{5}-\dfrac{1}{5}y+\dfrac{3}{5}
,$ 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 result to
\begin{array}{l}\require{cancel}
1-\dfrac{2}{3}y=\dfrac{9}{5}-\dfrac{1}{5}y+\dfrac{3}{5}
\\\\
15\left( 1-\dfrac{2}{3}y \right) =15\left( \dfrac{9}{5}-\dfrac{1}{5}y+\dfrac{3}{5} \right)
\\\\
15-10y=27-3y+9
\\\\
-10y+3y=27+9-15
\\\\
-7y=21
\\\\
y=\dfrac{21}{-7}
\\\\
y=-3
.\end{array}
Checking: If $y=-3,$ then
\begin{array}{l}\require{cancel}
1-\dfrac{2}{3}y=\dfrac{9}{5}-\dfrac{1}{5}y+\dfrac{3}{5}
\\\\
1-\dfrac{2}{3}(-3)=\dfrac{9}{5}-\dfrac{1}{5}(-3)+\dfrac{3}{5}
\\\\
1+2=\dfrac{9}{5}+\dfrac{3}{5}+\dfrac{3}{5}
\\\\
3=\dfrac{15}{5}
\\\\
3=3
\text{ (TRUE) }
.\end{array}
Hence, the solution is $
y=-3
.$