Question 265318
I'll help you graph {{{3x+7y=7}}}



{{{3x+7y=7}}} Start with the given equation.



{{{7y=7-3x}}} Subtract {{{3x}}} from both sides.



{{{7y=-3x+7}}} Rearrange the terms.



{{{y=(-3x+7)/(7)}}} Divide both sides by {{{7}}} to isolate y.



{{{y=((-3)/(7))x+(7)/(7)}}} Break up the fraction.



{{{y=-(3/7)x+1}}} Reduce.





Looking at {{{y=-(3/7)x+1}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=-3/7}}} and the y-intercept is {{{b=1}}} 



Since {{{b=1}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,1\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,1\right)]


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



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

{{{slope=rise/run}}}


Also, because the slope is {{{-3/7}}}, this means:


{{{rise/run=-3/7}}}



which shows us that the rise is -3 and the run is 7. This means that to go from point to point, we can go down 3  and over 7




So starting at *[Tex \LARGE \left(0,1\right)], go down 3 units 

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


and to the right 7 units to get to the next point *[Tex \LARGE \left(7,-2\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,1,.1)),
  blue(circle(0,1,.12)),
  blue(circle(0,1,.15)),
  blue(circle(7,-2,.15,1.5)),
  blue(circle(7,-2,.1,1.5)),
  blue(arc(0,1+(-3/2),2,-3,90,270)),
  blue(arc((7/2),-2,7,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-(3/7)x+1}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-(3/7)x+1),
  blue(circle(0,1,.1)),
  blue(circle(0,1,.12)),
  blue(circle(0,1,.15)),
  blue(circle(7,-2,.15,1.5)),
  blue(circle(7,-2,.1,1.5)),
  blue(arc(0,1+(-3/2),2,-3,90,270)),
  blue(arc((7/2),-2,7,2, 0,180))
)}}} So this is the graph of {{{y=-(3/7)x+1}}} through the points *[Tex \LARGE \left(0,1\right)] and *[Tex \LARGE \left(7,-2\right)]



I'll leave it to you to graph {{{3x+7y>7}}} (ie apply the proper shading)