91Ó°ÊÓ

The given linear system of equations is as follows:

$\left\{\begin{array}{l}y=\frac{2}{3}x-2\\ y=-\frac{1}{3}x-5\end{array}\right.$

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$y=\frac{2}{3}x-2$.

The equation is written in slope-inercept form.

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

Assume the second equation$y=-\frac{1}{3}x-5$.

The equation is written in slope-inercept form.

So, the slope $m$is $-\frac{1}{3}$and the $y$-intercept $b$is $-5$.

The graph of the two given lines is shown below:

From the graph, observe that the point of intersection is $(-3,-4)$.

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 $-4$for $y$into $y=\frac{2}{3}x-2$.

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

Substitute $-3$for $x$and $-4$for $y$into $y=-\frac{1}{3}x-5$.

$\begin{array}{rcl}-4& =& -\frac{1}{3}(-3)-5\\ -4& =& -1(-1)-5\\ -4& =& 1-5\\ -4& =& -4\left(\text{True}\right)\end{array}$

Conclude that the point $(-3,-4)$made both equations true.

Hence, $(-3,-4)$is a solution.