Question 365336
{{{3x-y=3}}} Start with the given equation.



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



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



{{{y=(-3x+3)/(-1)}}} Divide both sides by {{{-1}}} to isolate y.



{{{y=((-3)/(-1))x+(3)/(-1)}}} Break up the fraction.



{{{y=3x-3}}} Reduce.



Looking at {{{y=3x-3}}} 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=-3}}} 



Since {{{b=-3}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-3\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,-3\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-3,.1)),
  blue(circle(0,-3,.12)),
  blue(circle(0,-3,.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,-3\right)], go up 3 units 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-3,.1)),
  blue(circle(0,-3,.12)),
  blue(circle(0,-3,.15)),
  blue(arc(0,-3+(3/2),2,3,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,0\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-3,.1)),
  blue(circle(0,-3,.12)),
  blue(circle(0,-3,.15)),
  blue(circle(1,0,.15,1.5)),
  blue(circle(1,0,.1,1.5)),
  blue(arc(0,-3+(3/2),2,3,90,270)),
  blue(arc((1/2),0,1,2, 180,360))
)}}}



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


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