Solve each system of equations by graphing.

$\begin{array}{c}23.\frac{1}{2}x-y=0\\ \frac{1}{4}x+\frac{1}{2}y=-2\end{array}$

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 $\frac{1}{2}x-y=0$.

Rewrite the equation in form of slope-intercept form.

Add the term y on both the sides.

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 0.

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

Rewrite the equation in form of slope-intercept form.

Subtract both sides by $\frac{1}{4}x$.

Next, multiply both sides by 2 and rearrange the terms.

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 $\frac{1}{2}x-y=0$and blue line denotes the equation $\frac{1}{4}x+\frac{1}{2}y=-2$.

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 $\frac{1}{2}x-y=0$.

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

Since, this is true so the point $\left(-4,-2\right)$satisfy the equation $\frac{1}{2}x-y=0$.

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

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

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

Hence, the solution of the system of equations is .