Use elimination to solve each system of equations.

$\begin{array}{l}\mathbf{8}\mathbf{x}\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{18}\\ \mathbf{6}\mathbf{x}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{6}\end{array}$

The elimination method is the method in which addition or subtraction method is used to get an equation in the one variable.

If the coefficients are opposite, use the addition and when the coefficients are same, use the subtraction.

The given system of equation is:

$8x-5y=18$ … (1)

$6x+6y=-6$ … (2)

Multiply equation (1) by 6 and equation (2) by 5.

$\begin{array}{l}6\times (8x-5y)=6\left(18\right)\\ 48x-30y=108\left(3\right)\end{array}$

$\begin{array}{l}5\times (6x+6y)=5(-6)\\ 30x+30y=-30\left(4\right)\end{array}$

Note that both the equations (3) and (4) have $30y$in opposite signs, therefore, variable $y$can be eliminated if both equations are added.

$\begin{array}{l}48x-30y=108\\ \underset{¯}{(+)30x+30y=-30}\\ 78x=78\text{Dividebothsidesby78}\\ x=1\end{array}$

Substitute 1 for $x$into the equation (2) and solve for $y$.

$\begin{array}{c}6\left(1\right)+6y=-6\\ 6+6y-6=-6-6\\ 6y=-12\\ y=-2\end{array}$

The solution is $(1,-2)$.