Answer
$\left( 10,-4 \right)$
Work Step by Step
Using properties of equality, the second equation, $
x+3y=-2
,$ is equivalent to
\begin{array}{l}\require{cancel}
x+3y=-2
\\\\
x=-3y-2
.\end{array}
Substituting $x=-3y-2$ into the first equation results to
\begin{array}{l}\require{cancel}
\dfrac{2}{5}x+\dfrac{3}{4}y=1
\\\\
\dfrac{2}{5}(-3y-2)+\dfrac{3}{4}y=1
\\\\
20\left[\dfrac{2}{5}(-3y-2)+\dfrac{3}{4}y\right]=(1)20
\\\\
8(-3y-2)+15y=20
\\\\
-24y-16+15y=20
\\\\
-24y+15y=20+16
\\\\
-9y=36
\\\\
y=\dfrac{36}{-9}
\\\\
y=-4
.\end{array}
Substituting $y=-4$ into the second equation results to
\begin{array}{l}\require{cancel}
x+3y=-2
\\\\
x+3(-4)=-2
\\\\
x-12=-2
\\\\
x=-2+12
\\\\
x=10
.\end{array}
Hence, the solution set is $
\left( 10,-4 \right)
.$
