Question 864621
One way is to first isolate y



{{{3x + 2y = 8 }}}



{{{2y = 8-3x }}}



{{{2y = -3x+8 }}}



{{{y = (-3x+8)/2 }}}



{{{y = (-3x)/2+8/2 }}}



{{{y = (-3/2)x+4 }}}



Looking at {{{y=-(3/2)x+4}}} 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=4}}} 



Since {{{b=4}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,4\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,4\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,4,.1)),
  blue(circle(0,4,.12)),
  blue(circle(0,4,.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(0,4\right)], go down 3 units 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,4,.1)),
  blue(circle(0,4,.12)),
  blue(circle(0,4,.15)),
  blue(arc(0,4+(-3/2),2,-3,90,270))
)}}}


and to the right 2 units to get to the next point *[Tex \LARGE \left(2,1\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,4,.1)),
  blue(circle(0,4,.12)),
  blue(circle(0,4,.15)),
  blue(circle(2,1,.15,1.5)),
  blue(circle(2,1,.1,1.5)),
  blue(arc(0,4+(-3/2),2,-3,90,270)),
  blue(arc((2/2),1,2,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-(3/2)x+4}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,0,-(3/2)x+4),
  blue(circle(0,4,.1)),
  blue(circle(0,4,.12)),
  blue(circle(0,4,.15)),
  blue(circle(2,1,.15,1.5)),
  blue(circle(2,1,.1,1.5)),
  blue(arc(0,4+(-3/2),2,-3,90,270)),
  blue(arc((2/2),1,2,2, 0,180))
)}}} So this is the graph of {{{y=-(3/2)x+4}}} through the points *[Tex \LARGE \left(0,4\right)] and *[Tex \LARGE \left(2,1\right)]