Solve a system of linear equations by graphing.

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

The given linear system of equations is as follows:

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

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 $y=x-2$.

The equation is written in slope-inercept form.

So, the slope $m$is $1$and the $y$-intercept $b$is localid="1647835702969" $-2$.

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

The equation is written in slope-inercept form.

So, the slope $m$is $-3$and the $y$-intercept $b$is $2$.

The graph of the two given lines is shown below:

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

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 $1$for $x$and $-1$for $y$into $y=x-2$.

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

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

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

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

Hence, $(1,-1)$is a solution.