Solve a system of linear equations by graphing.

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

The given linear system of equations is as follows:

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

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-intercept form.

So, the slope $m$is $1$and the $y$-intercept $b$is $2$.

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

The equation is written in slope-intercept form.

So, the slope $m$is $-2$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 $(0,2)$.

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 $0$for $x$and $2$for $y$into $y=x+2$.

$\begin{array}{rcl}2& =& 0+2\\ 2& =& 2\left(\text{True}\right)\end{array}$

Substitute $0$for $x$and $2$for $y$into $y=-2x+2$.

$\begin{array}{rcl}2& =& -2\left(0\right)+2\\ 2& =& 0+2\\ 2& =& 2\left(\text{True}\right)\end{array}$

Conclude that the point $(0,2)$made both equations true.

Hence, $(0,2)$is a solution.