Solve each system of equations by graphing.

$\begin{array}{c}\frac{2}{3}x+y=-3\\ y-\frac{1}{3}x=6\end{array}$

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 $\frac{2}{3}x+y=-3$.

Rewrite the equation in form of slope-intercept form.

Subtract both sides by $\frac{2}{3}x$.

$\begin{array}{c}\frac{2}{3}x+y-\frac{2}{3}x=-3-\frac{2}{3}x\\ y=-3-\frac{2}{3}x\\ y=-\frac{2}{3}x-3\end{array}$

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

Now, consider the second equation $y-\frac{1}{3}x=6$.

Rewrite the equation in form of slope-intercept form.

Add the term $\frac{1}{3}x$on both the sides.

$\begin{array}{c}y-\frac{1}{3}x+\frac{1}{3}x=6+\frac{1}{3}x\\ y=6+\frac{1}{3}x\\ y=\frac{1}{3}x+6\end{array}$

Now, the equation is in the form $y=mx+c$. Here slope m of the line is $\frac{1}{3}$ 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 $\frac{2}{3}x+y=-3$ and blue line denotes the equation $y-\frac{1}{3}x=6$.

Therefore, the point of intersection is $\left(-9,3\right)$.

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

Substitute the point $\left(-9,3\right)$in the equation $\frac{2}{3}x+y=-3$.

Substitute x as $-9$ and y as 3 and check whether right hand side is equal to left hand side of the equation.

Since, this is true so the point $\left(-9,3\right)$satisfy the equation $\frac{2}{3}x+y=-3$.

Substitute the point $\left(-9,3\right)$in the equation $y-\frac{1}{3}x=6$.

Substitute x as $-9$ and y as 3 and check whether right hand side is equal to left hand side of the equation.

Since, this is true so the point $\left(-9,3\right)$satisfy the equation $y-\frac{1}{3}x=6$.

Hence, the solution of the system of equations is $\left(-9,3\right)$.