Solve a system of linear equations by graphing.

$\left\{\begin{array}{l}3x+y=-3\\ 2x+3y=5\end{array}\right.$

The given linear system of equations is as follows:

$\left\{\begin{array}{l}3x+y=-3\\ 2x+3y=5\end{array}\right.$

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 $3x+y=-3$.

$y=-3x-3$

So, the slope $m$is $-3$and the $y$-intercept $b$is$-3$.

Assume the second equation $2x+3y=5$.

$\begin{array}{rcl}3y& =& -2x+5\\ y& =& -\frac{2}{3}x+\frac{5}{3}\end{array}$

So, the slope $m$is $-\frac{2}{3}$and the $y$-intercept $b$is $\frac{5}{3}$.

The graph of the two given lines is shown below:

From the graph, observe that the point of intersection is $(-2,3)$.

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

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

Substitute $-2$for $x$and $3$for $y$into $2x+3y=5$.

$\begin{array}{rcl}2(-2)+3\left(3\right)& =& 5\\ -4+9& =& 5\\ 5& =& 5\left(\text{True}\right)\end{array}$

Conclude that the point localid="1647603924857" $(-2,3)$made both equations true.

Hence, localid="1647603933406" $(-2,3)$is a solution.