Solve each system of equations by graphing.

$\begin{array}{c}\frac{1}{4}x+2y=5\\ 2x-y=6\end{array}$

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 $\frac{1}{4}x+2y=5$.

Rewrite the equation in form of slope-intercept form.

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

Next, divide 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 $-\frac{1}{8}$ and intercept of y-axis c is $\frac{5}{2}$.

Now, consider the second equation $2x-y=6$.

Rewrite the equation in form of slope-intercept form.

Subtract both sides by $2x$.

Next, divide both sides by $-1$and rearrange the terms.

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 $-6$.

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 $\frac{1}{4}x+2y=5$ and blue line denotes the equation $2x-y=6$.

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 $\frac{1}{4}x+2y=5$.

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 $\frac{1}{4}x+2y=5$.

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.

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

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