Question 161943

{{{2y-3x=5}}} Start with the given equation.



{{{2y=5+3x}}} Add {{{3x}}} to both sides.



{{{2y=3x+5}}} Rearrange the terms.



{{{y=(3x+5)/(2)}}} Divide both sides by {{{2}}} to isolate y.



{{{y=(3/2)x+(5)/(2)}}} Break up the fraction.



{{{y=(3/2)x+5/2}}} Reduce.



So the equation {{{y=(3/2)x+5/2}}} is now in slope intercept form {{{y=mx+b}}} where the slope (ie gradient) is {{{m=3/2}}} and the y-intercept is {{{b=5/2}}} 





Since {{{b=5/2}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,\frac{5}{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,\frac{5}{2}\right)] which is (0,2.5)


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,5/2,.1)),
  blue(circle(0,5/2,.12)),
  blue(circle(0,5/2,.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 up 3  and over 2




So starting at *[Tex \LARGE \left(0,\frac{5}{2}\right)], go up 3 units 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,5/2,.1)),
  blue(circle(0,5/2,.12)),
  blue(circle(0,5/2,.15)),
  blue(arc(0,5/2+(3/2),2,3,90,270))
)}}}


and to the right 2 units to get to the next point *[Tex \LARGE \left(2,\frac{11}{2}\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,5/2,.1)),
  blue(circle(0,5/2,.12)),
  blue(circle(0,5/2,.15)),
  blue(circle(2,11/2,.15,1.5)),
  blue(circle(2,11/2,.1,1.5)),
  blue(arc(0,5/2+(3/2),2,3,90,270)),
  blue(arc((2/2),11/2,2,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=(3/2)x+5/2}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,(3/2)x+5/2),
  blue(circle(0,5/2,.1)),
  blue(circle(0,5/2,.12)),
  blue(circle(0,5/2,.15)),
  blue(circle(2,11/2,.15,1.5)),
  blue(circle(2,11/2,.1,1.5)),
  blue(arc(0,5/2+(3/2),2,3,90,270)),
  blue(arc((2/2),11/2,2,2, 180,360))
)}}} So this is the graph of {{{y=(3/2)x+5/2}}} through the points *[Tex \LARGE \left(0,\frac{5}{2}\right)] and *[Tex \LARGE \left(2,\frac{11}{2}\right)]