Solve a system of linear equations by graphing.

$\left\{\begin{array}{l}-2x+3y=-3\\ x+y=4\end{array}\right.$

The given linear system of equations is as follows:

$\left\{\begin{array}{l}-2x+3y=-3\\ x+y=4\end{array}\right.$

The standard form of slope-intercept form is given by $y=mx+b$, where $m$denotes the slope and $b$denotes the $y$-intercept.

Assume the first equation$-2x+3y=-3$.

$\begin{array}{rcl}3y& =& 2x-3\\ y& =& \frac{2}{3}x-1\end{array}$

So, the slope $m$is $\frac{2}{3}$and the $y$-intercept $b$is localid="1647606881409" $-1$.

Assume the second equation$x+y=4$.

$y=-x+4$

So, the slope $m$is $-1$and the $y$-intercept $b$is $4$.

The graph of the two given lines is shown below:

From the graph, observe that the point of intersection is $(3,1)$.

Substitute the values of the variables into each equation to check whether a point is a solution to a given system of two equations.

If the point makes both equations true, it is a solution to the given system.

Substitute $3$for $x$and $1$for $y$into $-2x+3y=-3$.

$\begin{array}{rcl}-2\left(3\right)+3\left(1\right)& =& -3\\ -6+3& =& -3\\ -3& =& -3\left(\text{True}\right)\end{array}$

Substitute $3$for $x$and $1$for $y$into $x+y=4$.

$\begin{array}{rcl}3+1& =& 4\\ 4& =& 4\left(\text{True}\right)\end{array}$

Conclude that the point $(3,1)$made both equations true.

Hence, localid="1647606976877" $(3,1)$is a solution.