Question 145602
# 1




Looking at {{{y=-(1/3)x+4}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-1/3}}} 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 {{{-1/3}}}, this means:


{{{rise/run=-1/3}}}



which shows us that the rise is -1 and the run is 3. This means that to go from point to point, we can go down 1  and over 3




So starting at *[Tex \LARGE \left(0,4\right)], go down 1 unit 

{{{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+(-1/2),2,-1,90,270))
)}}}


and to the right 3 units to get to the next point *[Tex \LARGE \left(3,3\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(3,3,.15,1.5)),
  blue(circle(3,3,.1,1.5)),
  blue(arc(0,4+(-1/2),2,-1,90,270)),
  blue(arc((3/2),3,3,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-(1/3)x+4}}}


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





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# 2


{{{3y-9x=30}}} Start with the given equation




{{{3y=30+9x}}} Add {{{9 x}}} to both sides



{{{3y=+9x+30}}} Rearrange the equation



{{{y=(+9x+30)/(3)}}} Divide both sides by {{{3}}}



{{{y=(+9/3)x+(30)/(3)}}} Break up the fraction



{{{y=3x+10}}} Reduce






Looking at {{{y=3x+10}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=3}}} and the y-intercept is {{{b=10}}} 



Since {{{b=10}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,10\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,10\right)]


{{{drawing(500,500,-10,10,-5,15,
  grid(1),
  blue(circle(0,10,.1)),
  blue(circle(0,10,.12)),
  blue(circle(0,10,.15))
)}}}



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

{{{slope=rise/run}}}


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


{{{rise/run=3/1}}}



which shows us that the rise is 3 and the run is 1. This means that to go from point to point, we can go up 3  and over 1




So starting at *[Tex \LARGE \left(0,10\right)], go up 3 units 

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


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,13\right)]

{{{drawing(500,500,-10,10,-5,15,
  grid(1),
  blue(circle(0,10,.1)),
  blue(circle(0,10,.12)),
  blue(circle(0,10,.15)),
  blue(circle(1,13,.15,1.5)),
  blue(circle(1,13,.1,1.5)),
  blue(arc(0,10+(3/2),2,3,90,270)),
  blue(arc((1/2),13,1,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=3x+10}}}


{{{drawing(500,500,-10,10,-5,15,
  grid(1),
  graph(500,500,-10,10,-5,15,3x+10),
  blue(circle(0,10,.1)),
  blue(circle(0,10,.12)),
  blue(circle(0,10,.15)),
  blue(circle(1,13,.15,1.5)),
  blue(circle(1,13,.1,1.5)),
  blue(arc(0,10+(3/2),2,3,90,270)),
  blue(arc((1/2),13,1,2, 180,360))
)}}} So this is the graph of {{{y=3x+10}}} through the points *[Tex \LARGE \left(0,10\right)] and *[Tex \LARGE \left(1,13\right)]