Question 201975
Step 1: Graph {{{x - y = 10}}}



{{{x-y=10}}} Start with the given equation.



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



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



{{{y=(-x+10)/(-1)}}} Divide both sides by {{{-1}}} to isolate y.



{{{y=(-x)/(-1)+(10)/(-1)}}} Break up the fraction.



{{{y=x-10}}} Reduce.



Looking at {{{y=x-10}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=1}}} and the y-intercept is {{{b=-10}}} 



Since {{{b=-10}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-10\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,-10\right)]


{{{drawing(500,500,-10,10,-15,5,
  grid(1),
  blue(circle(0,-10,.1)),
  blue(circle(0,-10,.12)),
  blue(circle(0,-10,.15))
)}}}



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

{{{slope=rise/run}}}


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


{{{rise/run=1/1}}}



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




So starting at *[Tex \LARGE \left(0,-10\right)], go up 1 unit 

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


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,-9\right)]

{{{drawing(500,500,-10,10,-15,5,
  grid(1),
  blue(circle(0,-10,.1)),
  blue(circle(0,-10,.12)),
  blue(circle(0,-10,.15)),
  blue(circle(1,-9,.15,1.5)),
  blue(circle(1,-9,.1,1.5)),
  blue(arc(0,-10+(1/2),2,1,90,270)),
  blue(arc((1/2),-9,1,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=x-10}}}


{{{drawing(500,500,-10,10,-15,5,
  grid(1),
  graph(500,500,-10,10,-15,5,x-10),
  blue(circle(0,-10,.1)),
  blue(circle(0,-10,.12)),
  blue(circle(0,-10,.15)),
  blue(circle(1,-9,.15,1.5)),
  blue(circle(1,-9,.1,1.5)),
  blue(arc(0,-10+(1/2),2,1,90,270)),
  blue(arc((1/2),-9,1,2, 180,360))
)}}} So this is the graph of {{{y=x-10}}} (and {{{x-y=10}}}) through the points *[Tex \LARGE \left(0,-10\right)] and *[Tex \LARGE \left(1,-9\right)]



------------------------------------------------------------------


Step 2: Graph {{{x - y = -2}}}



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



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



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



{{{y=(-x-2)/(-1)}}} Divide both sides by {{{-1}}} to isolate y.



{{{y=(-x)/(-1)+(-2)/(-1)}}} Break up the fraction.



{{{y=x+2}}} Reduce.




Looking at {{{y=x+2}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=1}}} 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,-15,5,
  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 {{{1}}}, this means:


{{{rise/run=1/1}}}



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




So starting at *[Tex \LARGE \left(0,2\right)], go up 1 unit 

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


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,3\right)]

{{{drawing(500,500,-10,10,-15,5,
  grid(1),
  blue(circle(0,2,.1)),
  blue(circle(0,2,.12)),
  blue(circle(0,2,.15)),
  blue(circle(1,3,.15,1.5)),
  blue(circle(1,3,.1,1.5)),
  blue(arc(0,2+(1/2),2,1,90,270)),
  blue(arc((1/2),3,1,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=x+2}}}


{{{drawing(500,500,-10,10,-15,5,
  grid(1),
  graph(500,500,-10,10,-15,5,x+2),
  blue(circle(0,2,.1)),
  blue(circle(0,2,.12)),
  blue(circle(0,2,.15)),
  blue(circle(1,3,.15,1.5)),
  blue(circle(1,3,.1,1.5)),
  blue(arc(0,2+(1/2),2,1,90,270)),
  blue(arc((1/2),3,1,2, 180,360))
)}}} So this is the graph of {{{y=x+2}}} (and {{{x-y=-2}}}) through the points *[Tex \LARGE \left(0,2\right)] and *[Tex \LARGE \left(1,3\right)]


---------------------------------------------------------------


Step 3: Graph the two equations together:


{{{ drawing(500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,(10-x)/(-1),(-2-x)/(-1))

)}}}



Graph of {{{x-y=10}}} (red) and {{{x-y=-2}}} (green)



From the graph, we can see that the two lines are parallel. So they will NEVER cross. 



This means that there are no solutions.



So the system is inconsistent.