Solve each system of equations by graphing.

$\begin{array}{c}2x-y=6\\ -x+8y=12\end{array}$

Equation of line in slope intercept form is expressed below.

$y=mx+c$

Where m is the slope and c is the intercept of y-axis.

Consider the first equation $2x-y=6$.

Rewrite the equation in slope-intercept form.

Subtract $2x$from both the sides.

$\begin{array}{c}2x-y-2x=6-2x\\ -y=6-2x\end{array}$

Multiply both sides by $-1$

$y=2x-6$

The equation is in the form $y=mx+c$. Here slope m of the line is 2 and intercept of y-axis c is $-6$.

Now, consider the second equation $-x+8y=12$.

Rewrite the equation in form of slope-intercept form.

Add x on both the sides.

$\begin{array}{c}-x+8y+x=12+x\\ 8y=12+x\\ 8y=x+12\end{array}$

Divide both sides by 8.

$y=\frac{1}{8}x+\frac{3}{2}$

Now, the equation is in the form $y=mx+c$. Here slope m of the line is $\frac{1}{8}$ and intercept of y-axis c is $\frac{3}{2}$.

Plot the equations on the same plane and the point where both the equations intersect is the solution of the system of the equations.

The red line denotes the equation $2x-y=6$ and blue line denotes the equation $-x+8y=12$.

Therefore, the point of intersection is $\left(4,2\right)$.

The point of intersection is solution of system of equations if the point satisfies both the equation.

Substitute the point $\left(4,2\right)$ in the equation $2x-y=6$.

Substitute x as 4 and y as 2 and check whether right hand side is equal to left hand side of the equation.

$\begin{array}{c}2\left(4\right)-2=6\\ 8-2=6\\ 6=6\end{array}$

Since, this is true so the point $\left(4,2\right)$ satisfy the equation $2x-y=6$.

Substitute the point $\left(4,2\right)$ in the equation $-x+8y=12$.

Substitute x as 4 and y as 2 and check whether right hand side is equal to left hand side of the equation.

$\begin{array}{c}-4+8\left(2\right)=12\\ -4+16=12\\ 12=12\end{array}$

Since, this is true so the point $\left(4,2\right)$ satisfy the equation $-x+8y=12$.

Hence, the solution of the system of equations is $\left(4,2\right)$.