Question 341666


{{{4x+3y=9}}} Start with the given equation.



{{{3y=9-4x}}} Subtract {{{4x}}} from both sides.



{{{3y=-4x+9}}} Rearrange the terms.



{{{y=(-4x+9)/(3)}}} Divide both sides by {{{3}}} to isolate y.



{{{y=((-4)/(3))x+(9)/(3)}}} Break up the fraction.



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



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


{{{rise/run=-4/3}}}



which shows us that the rise is -4 and the run is 3. This means that to go from point to point, we can go down 4  and over 3




So starting at *[Tex \LARGE \left(0,3\right)], go down 4 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+(-4/2),2,-4,90,270))
)}}}


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



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


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



If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com?Subject=Algebra%20Help">jim_thompson5910@hotmail.com</a>


Also, feel free to check out my <a href="http://www.freewebs.com/jimthompson5910/home.html">tutoring website</a>


Jim