Question 127329


{{{3x+2y=12}}} Start with the given equation



{{{2y=12-3x}}}  Subtract {{{3 x}}} from both sides



{{{2y=-3x+12}}} Rearrange the equation



{{{y=(-3x+12)/(2)}}} Divide both sides by {{{2}}}



{{{y=(-3/2)x+(12)/(2)}}} Break up the fraction



{{{y=(-3/2)x+6}}} Reduce




So the equation is now in slope-intercept form ({{{y=mx+b}}}) where the slope is {{{m=-3/2}}} and the y-intercept is {{{b=6}}}





Looking at {{{y=-(3/2)x+6}}} 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=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 {{{-3/2}}}, 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(0,6\right)], go down 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 2 units to get to the next point *[Tex \LARGE \left(2,3\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(2,3,.15,1.5)),
  blue(circle(2,3,.1,1.5)),
  blue(arc(0,6+(-3/2),2,-3,90,270)),
  blue(arc((2/2),3,2,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-(3/2)x+6}}}


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