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

$y=-3$
$\bf{\text{Solution Outline:}}$ 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 $y=-3 .$