Question 208161
There are two ways to do this (actually there are more, but I'm only showing two here).



Method #1



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



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



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



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



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



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





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


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



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




So starting at *[Tex \LARGE \left(0,3\right)], go down 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 2 units to get to the next point *[Tex \LARGE \left(2,0\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(2,0,.15,1.5)),
  blue(circle(2,0,.1,1.5)),
  blue(arc(0,3+(-3/2),2,-3,90,270)),
  blue(arc((2/2),0,2,2, 0,180))
)}}}



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


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





<hr>



OR....



Method #2



To plot {{{3x + 2y = 6}}}, we can use the x and y intercepts.




<h4>x-intercept</h4>

To find the x-intercept, plug in {{{y=0}}} and solve for x



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



{{{3x + 2(0) = 6}}} Plug in {{{y=0}}}.



{{{3x + 0 = 6}}} Multiply {{{2}}} and 0 to get 0.



{{{3x + 0 = 6}}} Simplify.



{{{3x=6}}} Multiply.



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



{{{x=2}}} Reduce.



So the x-intercept is *[Tex \LARGE \left(2,0\right)].



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



<h4>y-intercept</h4>

To find the y-intercept, plug in {{{x=0}}} and solve for y



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



{{{3(0) + 2y = 6}}} Plug in {{{x=0}}}.



{{{0 + 2y = 6}}} Multiply {{{3}}} and 0 to get 0.



{{{ + 2y = 6}}} Simplify.



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



{{{y=3}}} Reduce.



So the y-intercept is *[Tex \LARGE \left(0,3\right)].



Now plot the x and y intercepts:


{{{ drawing(500, 500, -10, 10, -10, 10,
grid(0),
graph(500, 500, -10, 10, -10, 10,0),
circle(0,3,0.05),circle(0,3,0.08),circle(0,3,0.10),circle(0,3,0.12),circle(0,3,0.14),
circle(2,0,0.05),circle(2,0,0.08),circle(2,0,0.10),circle(2,0,0.12),circle(2,0,0.14)
)}}} 




Now draw a line through the points to graph the equation {{{3x + 2y = 6}}}



{{{ drawing(500, 500, -10, 10, -10, 10,
grid(0),
graph(500, 500, -10, 10, -10, 10,(6-3x)/(2)),
circle(0,3,0.05),circle(0,3,0.08),circle(0,3,0.10),circle(0,3,0.12),circle(0,3,0.14),
circle(2,0,0.05),circle(2,0,0.08),circle(2,0,0.10),circle(2,0,0.12),circle(2,0,0.14)
)}}} Graph of {{{3x + 2y = 6}}}