Solve each system of equations by graphing.

$\begin{array}{c}y=3x+10\\ y=-x+6\end{array}$

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 $y=3x+10$.

The equation is already in the form $y=mx+c$. Here slope m of the line is $3$ and intercept of y-axis c is $10$.

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

This equation is also in the form $y=mx+c$. Here slope m of the line is $- 1$ 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 $y=3x+10$and blue line denotes the equation $y=-x+6$.

Therefore, the point of intersection is $\left(- 1 , 7\right)$.

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

Substitute the point $\left(- 1 , 7\right)$in the equation $y=3x+10$.

Substitute x as $- 1$ and y as $7$and check whether right hand side is equal to left hand side of the equation.

$\begin{array}{c}7=3\left(-1\right)+10\\ 7=-3+10\\ 7=7\end{array}$

Since, this is true so the point $\left(- 1 , 7\right)$satisfy the equation $y=3x+10$.

Substitute the point $\left(- 1 , 7\right)$in the equation $y=-x+6$.

Substitute x as $- 1$ and y as $7$and check whether right hand side is equal to left hand side of the equation.

$\begin{array}{c}7=-\left(-1\right)+6\\ 7=1+6\\ 7=7\end{array}$

Since, this is true so the point $\left(- 1 , 7\right)$ satisfy the equation $y=-x+6$.

Hence, the solution of the system of equations is $\left(- 1 , 7\right)$.