Question 236206
I'm assuming you want to graph this.




{{{2x+8y=2}}} Start with the given equation.



{{{8y=2-2x}}} Subtract {{{2x}}} from both sides.



{{{8y=-2x+2}}} Rearrange the terms.



{{{y=(-2x+2)/(8)}}} Divide both sides by {{{8}}} to isolate y.



{{{y=((-2)/(8))x+(2)/(8)}}} Break up the fraction.



{{{y=-(1/4)x+1/4}}} Reduce.





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



Since {{{b=1/4}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,\frac{1}{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,\frac{1}{4}\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,1/4,.1)),
  blue(circle(0,1/4,.12)),
  blue(circle(0,1/4,.15))
)}}}



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

{{{slope=rise/run}}}


Also, because the slope is {{{-1/4}}}, this means:


{{{rise/run=-1/4}}}



which shows us that the rise is -1 and the run is 4. This means that to go from point to point, we can go down 1  and over 4




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

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


and to the right 4 units to get to the next point *[Tex \LARGE \left(4,-\frac{3}{4}\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,1/4,.1)),
  blue(circle(0,1/4,.12)),
  blue(circle(0,1/4,.15)),
  blue(circle(4,-3/4,.15,1.5)),
  blue(circle(4,-3/4,.1,1.5)),
  blue(arc(0,1/4+(-1/2),2,-1,90,270)),
  blue(arc((4/2),-3/4,4,2, 0,180))
)}}}



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


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