Question 723128
I believe the first equation was meant to be {{{y=-(2/3)x+2}}} so that is how I will solve it.
 
You are expected to find the answer from the graph, so pretend you do not know the answer.
 
{{{y=-(2/3)x+2}}} and {{{y=-2x-2}}} are linear equations that graph as straight lines, and are given in slope-intercept form. 
For {{{y=-(2/3)x+2}}} , the slope is {{{-(2/3)}}} and the intercept is {{{2}}}.
For {{{y=-2x-2}}} , the slope is {{{-2}}} and the intercept is {{{-2}}}.
When the equation is in the y= ... slope-intercept from, the number multiplying the x is the slope and the number added at the end is the intercept, or y-intercept (the y-coordinate of the point where the line crosses the y-axis).
 
We plot the y- intercepts first. Then we use the slope to go from that point to another point,and repeat the procedure as needed/desired to help us draw the line. No calculations, just counting.
 
Graphing {{{y=-2x-2}}} :
A slope of {{{-2}}} means that for every 1 unit increase in x, y increases by -2 (meaning that it decreases by 2 units).
So from (0,-2), the intercept point for {{{y=-2x-2}}} , we move one unit right and 2 down to mark the next point. We repeat that procedure for the next and the next points. Finally we draw a line through our points.
{{{drawing(300,300,-6,6,-7,5,
grid(1),circle(0,-2,0.2),
blue(circle(1,-4,0.2)),blue(circle(2,-6,0.2)),
blue(line(-5,8,5,-12))
)}}}
 
Adding the line for  {{{y=-(2/3)x+2}}} to the graph:
A slope of {{{-(2/3)}}} means that the ratio of increase in to to increase in x is {{{-(2/3)}}} , meaning that for every 3 unit increase in x, y increases by -2 (it decreases by 2 units).
So from (0,2), the intercept point for {{{y=-(2/3)x+2}}} , we move 3 unit right and 2 down to mark the next point. Those two points are far enough apart to allow us to draw the line accurately, so we draw a line through our points.
{{{drawing(300,300,-6,6,-7,5,
grid(1),circle(0,-2,0.2),
blue(circle(1,-4,0.2)),blue(circle(2,-6,0.2)),
blue(line(-5,8,5,-12)),
circle(0,2,0.2),red(circle(3,0,0.2)),
red(line(-6,6,6,-2))
)}}}
The two lines seem to intersect at point (-3,4), corresponding to {{{highlight(x=-3)}}} and {{{highlight(y=4)}}} .
We must verify, because drawings can be inaccurate.
If that pair of values satisfies both equations, it is the solution of the system, found by graphing.
Substituting {{{x=-3}}} into {{{y=-(2/3)x+2}}} we get
{{{y=-(2/3)(-3)+2}}} --> {{{y=2+2}}} --> {{{y=4}}}
Substituting {{{x=-3}}} into {{{y=-2x-2}}} we get
{{{y=-2(-3)-2}}} --> {{{y=6-2}}} --> {{{y=4}}}
THE SOLUTION IS VERIFIED.