Question 490739
You start by graphing each equation separately, then you combine them to form one graph.



So let's start with graphing y=3x-6




Looking at {{{y=3x-6}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=3}}} 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 {{{3}}}, this means:


{{{rise/run=3/1}}}



which shows us that the rise is 3 and the run is 1. This means that to go from point to point, we can go up 3  and over 1




So starting at *[Tex \LARGE \left(0,-6\right)], go up 3 units 

{{{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+(3/2),2,3,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,-3\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,-3,.15,1.5)),
  blue(circle(1,-3,.1,1.5)),
  blue(arc(0,-6+(3/2),2,3,90,270)),
  blue(arc((1/2),-3,1,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=3x-6}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,3x-6),
  blue(circle(0,-6,.1)),
  blue(circle(0,-6,.12)),
  blue(circle(0,-6,.15)),
  blue(circle(1,-3,.15,1.5)),
  blue(circle(1,-3,.1,1.5)),
  blue(arc(0,-6+(3/2),2,3,90,270)),
  blue(arc((1/2),-3,1,2, 180,360))
)}}} So this is the graph of {{{y=3x-6}}} through the points *[Tex \LARGE \left(0,-6\right)] and *[Tex \LARGE \left(1,-3\right)]



Now use the same technique to graph y = -x+2


Once you've done that, and had a bit of practice with graphing, graph the two lines on the same grid. Where they intersect will give you the solution of the system.