Solve each system of equations by graphing.

$\begin{array}{c}y-2x=8\\ y=\frac{1}{2}x-4\end{array}$

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 $y-2x=8$.

Rewrite the equation in form of slope-intercept form.

Add both sides by $2x$.

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

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

Now, consider the second equation $y=\frac{1}{2}x-4$.

Rewrite the equation in form of slope-intercept form.

$y=\frac{1}{2}x+\left(-4\right)$

Now, the equation is in the form $y=mx+c$. Here slope m of the line is $\frac{1}{2}$ and intercept of y-axis c is $-4$.

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 $y-2x=8$ and blue line denotes the equation $y=\frac{1}{2}x-4$.

Therefore, the point of intersection is $\left(-8,-8\right)$.

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

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

Substitute x as $-8$ and y as $-8$ and check whether right hand side is equal to left hand side of the equation.

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

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

Substitute the point $\left(-8,-8\right)$ in the equation $y=\frac{1}{2}x-4$.

Substitute x as $-8$ and y as $-8$ and check whether right hand side is equal to left hand side of the equation.

$\begin{array}{c}-8=\frac{1}{2}\left(-8\right)-4\\ -8=-4-4\\ -8=-8\end{array}$

Since, this is true so the point $\left(-8,-8\right)$ satisfy the equation $y=\frac{1}{2}x-4$.

Hence, the solution of the system of equations is $\left(-8,-8\right)$.