Answer
$x=-1$ and $y=\dfrac{1}{2}$
Work Step by Step
To solve the given system,
\begin{align*}\require{cancel}
5x+6y&=-2
\\
3x+10y&=2
,\end{align*}
Cancel the $y$-variable by multiplying the first equation by $-5$ and the second equation by $3$. This results to the equivalent system
\begin{align*}\require{cancel}
&\left\{
\begin{array}{ll}
(-5)(5x+6y)&=(-2)(-5)
\\
(3)(3x+10y)&=(2)(3)
\end{array}
\right.
\\\\&
\left\{
\begin{array}{ll}
-25x-30y&=10
\\
9x+30y&=6
\end{array}
\right.
.\end{align*}
Adding the two equations above and using the properties of equality to solve for the remaining variable result in
\begin{align*}
-16x&=16
\\\\
\dfrac{\cancel{-16}x}{\cancel{-16}}&=\dfrac{16}{-16}
\\\\
x&=-1
.\end{align*}
Substituting $x=-1$ in the first given equation, $5x+6y=-2,$ results in
\begin{align*}
5(-1)+6y&=-2
\\
-5+6y&=-2
\\
-5+5+6y&=-2+5
\\
6y&=3
\\\\
\dfrac{\cancel6y}{\cancel6}&=\dfrac{3}{6}
\\\\
y&=\dfrac{1}{2}
.\end{align*}
Hence, the solution to the given system is $x=-1$ and $y=\dfrac{1}{2}$.
