Question 169796
In order to graph {{{2x+3y=6}}}, we first need to solve for y



{{{2x+3y=6}}} Start with the given equation.



{{{3y=6-2x}}} Subtract {{{2x}}} from both sides.



{{{3y=-2x+6}}} Rearrange the terms.



{{{y=(-2x+6)/(3)}}} Divide both sides by {{{3}}} to isolate y.



{{{y=((-2)/(3))x+(6)/(3)}}} Break up the fraction.



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





Looking at {{{y=-(2/3)x+2}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-2/3}}} and the y-intercept is {{{b=2}}} 



Since {{{b=2}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,2\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,2\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,2,.1)),
  blue(circle(0,2,.12)),
  blue(circle(0,2,.15))
)}}}



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

{{{slope=rise/run}}}


Also, because the slope is {{{-2/3}}}, this means:


{{{rise/run=-2/3}}}



which shows us that the rise is -2 and the run is 3. This means that to go from point to point, we can go down 2  and over 3




So starting at *[Tex \LARGE \left(0,2\right)], go down 2 units 

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


and to the right 3 units to get to the next point *[Tex \LARGE \left(3,0\right)]

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



Now draw a line through these points to graph {{{y=-(2/3)x+2}}}


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