Question 189578
First, let's graph {{{y=2x-5}}}




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



Since {{{b=-5}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-5\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,-5\right)]


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



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

{{{slope=rise/run}}}


Also, because the slope is {{{2}}}, this means:


{{{rise/run=2/1}}}



which shows us that the rise is 2 and the run is 1. This means that to go from point to point, we can go up 2  and over 1




So starting at *[Tex \LARGE \left(0,-5\right)], go up 2 units 

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


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

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



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


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




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


Now let's graph {{{y=x-6}}}






Looking at {{{y=x-6}}} 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=-6}}} 



Since {{{b=-6}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-6\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,-6\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-6,.1)),
  blue(circle(0,-6,.12)),
  blue(circle(0,-6,.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,-6\right)], go up 1 unit 

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


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,-5\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-6,.1)),
  blue(circle(0,-6,.12)),
  blue(circle(0,-6,.15)),
  blue(circle(1,-5,.15,1.5)),
  blue(circle(1,-5,.1,1.5)),
  blue(arc(0,-6+(1/2),2,1,90,270)),
  blue(arc((1/2),-5,1,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=x-6}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,x-6),
  blue(circle(0,-6,.1)),
  blue(circle(0,-6,.12)),
  blue(circle(0,-6,.15)),
  blue(circle(1,-5,.15,1.5)),
  blue(circle(1,-5,.1,1.5)),
  blue(arc(0,-6+(1/2),2,1,90,270)),
  blue(arc((1/2),-5,1,2, 180,360))
)}}} So this is the graph of {{{y=x-6}}} through the points *[Tex \LARGE \left(0,-6\right)] and *[Tex \LARGE \left(1,-5\right)]



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


Now let's graph the two equations on the same coordinate plane:


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,2x-5,x-6)
)}}} Graph of {{{y=2x-5}}} (red) and {{{y=x-6}}} (green)



From the graph, we can see that the two lines intersect at the point (-1,-7)



So the solutions are {{{x=-1}}} and {{{y=-7}}}. 



So the system is consistent and independent.