Question 174157


{{{5x+2y=-6}}} Start with the given equation.



{{{2y=-6-5x}}} Subtract {{{5x}}} from both sides.



{{{2y=-5x-6}}} Rearrange the terms.



{{{y=(-5x-6)/(2)}}} Divide both sides by {{{2}}} to isolate y.



{{{y=((-5)/(2))x+(-6)/(2)}}} Break up the fraction.



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





Looking at {{{y=-(5/2)x-3}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-5/2}}} 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 {{{-5/2}}}, this means:


{{{rise/run=-5/2}}}



which shows us that the rise is -5 and the run is 2. This means that to go from point to point, we can go down 5  and over 2




So starting at *[Tex \LARGE \left(0,-3\right)], go down 5 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+(-5/2),2,-5,90,270))
)}}}


and to the right 2 units to get to the next point *[Tex \LARGE \left(2,-8\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(2,-8,.15,1.5)),
  blue(circle(2,-8,.1,1.5)),
  blue(arc(0,-3+(-5/2),2,-5,90,270)),
  blue(arc((2/2),-8,2,2, 0,180))
)}}}



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


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-(5/2)x-3),
  blue(circle(0,-3,.1)),
  blue(circle(0,-3,.12)),
  blue(circle(0,-3,.15)),
  blue(circle(2,-8,.15,1.5)),
  blue(circle(2,-8,.1,1.5)),
  blue(arc(0,-3+(-5/2),2,-5,90,270)),
  blue(arc((2/2),-8,2,2, 0,180))
)}}} So this is the graph of {{{y=-(5/2)x-3}}} through the points *[Tex \LARGE \left(0,-3\right)] and *[Tex \LARGE \left(2,-8\right)]