Question 253245
First let's graph {{{2x-3y = -2}}}. To do so, we need to solve for 'y'



{{{2x-3y=-2}}} Start with the first equation.



{{{-3y=-2-2x}}} Subtract {{{2x}}} from both sides.



{{{-3y=-2x-2}}} Rearrange the terms.



{{{y=(-2x-2)/(-3)}}} Divide both sides by {{{-3}}} to isolate y.



{{{y=((-2)/(-3))x+(-2)/(-3)}}} Break up the fraction.



{{{y=(2/3)x+2/3}}} Reduce.



Now in order to graph {{{y=(2/3)x+2/3}}}, we need two points. 


Take note that when {{{x=2}}}, then {{{y=(2/3)(2)+2/3=4/3+2/3=6/3=2}}}. So we have one point (2,2). 



Also, when {{{x=5}}}, {{{y=(2/3)(5)+2/3=10/3+2/3=12/3=4}}} giving us another point (5,4). Plot these points to get



{{{drawing(500,500,-6.5,13.5,-6.5,13.5,
graph(500,500,-6.5,13.5,-6.5,13.5,0),
grid(1),
circle(2,6/3,0.1),
circle(2,6/3,0.12),
circle(2,6/3,0.15),
circle(5,12/3,0.1),
circle(5,12/3,0.12),
circle(5,12/3,0.15)
) }}}




Now draw a straight line through the two points. This line is the graph of {{{y=(2/3)x+2/3}}}


{{{drawing(500,500,-6.5,13.5,-6.5,13.5,
graph(500,500,-6.5,13.5,-6.5,13.5,(2/3)x+2/3),
grid(1),
circle(2,6/3,0.1),
circle(2,6/3,0.12),
circle(2,6/3,0.15),
circle(5,12/3,0.1),
circle(5,12/3,0.12),
circle(5,12/3,0.15)
) }}} 

Graph of {{{y=(2/3)x+2/3}}} through the two points (2,2) and (5,4)



-------------------------------------------------------------------------


Now let's graph {{{x+y = -6}}}



{{{x+y=-6}}} Start with the second equation.



{{{y=-6-x}}} Subtract {{{x}}} from both sides.



{{{y=-x-6}}} Rearrange the terms.



Looking at {{{y=-x-6}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-1}}} and the y-intercept is {{{b=-6}}} 



Since {{{b=-6}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-6\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,-6\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-6,.1)),
  blue(circle(0,-6,.12)),
  blue(circle(0,-6,.15))
)}}}



Now since the slope is comprised of the "rise" over the "run" this means

{{{slope=rise/run}}}


Also, because the slope is {{{-1}}}, this means:


{{{rise/run=-1/1}}}



which shows us that the rise is -1 and the run is 1. This means that to go from point to point, we can go down 1  and over 1




So starting at *[Tex \LARGE \left(0,-6\right)], go down 1 unit 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-6,.1)),
  blue(circle(0,-6,.12)),
  blue(circle(0,-6,.15)),
  blue(arc(0,-6+(-1/2),2,-1,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,-7\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-6,.1)),
  blue(circle(0,-6,.12)),
  blue(circle(0,-6,.15)),
  blue(circle(1,-7,.15,1.5)),
  blue(circle(1,-7,.1,1.5)),
  blue(arc(0,-6+(-1/2),2,-1,90,270)),
  blue(arc((1/2),-7,1,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-x-6}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-x-6),
  blue(circle(0,-6,.1)),
  blue(circle(0,-6,.12)),
  blue(circle(0,-6,.15)),
  blue(circle(1,-7,.15,1.5)),
  blue(circle(1,-7,.1,1.5)),
  blue(arc(0,-6+(-1/2),2,-1,90,270)),
  blue(arc((1/2),-7,1,2, 0,180))
)}}} 


So this is the graph of {{{y=-x-6}}} through the points *[Tex \LARGE \left(0,-6\right)] and *[Tex \LARGE \left(1,-7\right)]



------------------------------------------------------------------------


Now let's graph the two equations on the same coordinate plane


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,(2/3)x+2/3,-x-6)
)}}} 

Graph of {{{y=(2/3)x+2/3}}} (red) and {{{y=-x-6}}} (green)



From the graph, we see that the two lines intersect at the point (-4, -2). So the solutions are {{{x=-4}}} and {{{y=-2}}}



Check:


{{{2x-3y=-2}}} Start with the first equation.



{{{2(-4)-3(-2)=-2}}} Plug in {{{x=-4}}} and {{{y=-2}}}



{{{-8+6=-2}}} Multiply



{{{-2=-2}}} Add. Since the equation is true, this verifies our answer on the first equation.



-------------


{{{x+y=-6}}} Move onto the second equation.



{{{-4-2=-6}}} Plug in {{{x=-4}}} and {{{y=-2}}}



{{{-6=-6}}} Subtract. Since the equation is true, this verifies our answer for the second equation.



Since the values {{{x=-4}}} and {{{y=-2}}} satisfy both equations, this verifies our answer completely.