Question 254910
In order to graph {{{6x+5y=10}}}, we first must solve for y.



{{{6x+5y=10}}} Start with the given equation.



{{{5y=10-6x}}} Subtract {{{6x}}} from both sides.



{{{5y=-6x+10}}} Rearrange the terms.



{{{y=(-6x+10)/(5)}}} Divide both sides by {{{5}}} to isolate y.



{{{y=((-6)/(5))x+(10)/(5)}}} Break up the fraction.



{{{y=-(6/5)x+2}}} Reduce.



Looking at {{{y=-(6/5)x+2}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-6/5}}} and the y-intercept is {{{b=2}}} 



Since {{{b=2}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,2\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,2\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,2,.1)),
  blue(circle(0,2,.12)),
  blue(circle(0,2,.15))
)}}}



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

{{{slope=rise/run}}}


Also, because the slope is {{{-6/5}}}, this means:


{{{rise/run=-6/5}}}



which shows us that the rise is -6 and the run is 5. This means that to go from point to point, we can go down 6  and over 5




So starting at *[Tex \LARGE \left(0,2\right)], go down 6 units 

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,2,.1)),
  blue(circle(0,2,.12)),
  blue(circle(0,2,.15)),
  blue(arc(0,2+(-6/2),2,-6,90,270))
)}}}


and to the right 5 units to get to the next point *[Tex \LARGE \left(5,-4\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,2,.1)),
  blue(circle(0,2,.12)),
  blue(circle(0,2,.15)),
  blue(circle(5,-4,.15,1.5)),
  blue(circle(5,-4,.1,1.5)),
  blue(arc(0,2+(-6/2),2,-6,90,270)),
  blue(arc((5/2),-4,5,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-(6/5)x+2}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-(6/5)x+2),
  blue(circle(0,2,.1)),
  blue(circle(0,2,.12)),
  blue(circle(0,2,.15)),
  blue(circle(5,-4,.15,1.5)),
  blue(circle(5,-4,.1,1.5)),
  blue(arc(0,2+(-6/2),2,-6,90,270)),
  blue(arc((5/2),-4,5,2, 0,180))
)}}} So this is the graph of {{{y=-(6/5)x+2}}} through the points *[Tex \LARGE \left(0,2\right)] and *[Tex \LARGE \left(5,-4\right)]