Question 243642
First let's graph {{{y=(1/2)x+2}}}



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


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



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




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

{{{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+(1/2),2,1,90,270))
)}}}


and to the right 2 units to get to the next point *[Tex \LARGE \left(2,3\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(2,3,.15,1.5)),
  blue(circle(2,3,.1,1.5)),
  blue(arc(0,2+(1/2),2,1,90,270)),
  blue(arc((2/2),3,2,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=(1/2)x+2}}}


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



Now let's graph {{{y=(2/3)x-1}}}




Looking at {{{y=(2/3)x-1}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=2/3}}} 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 {{{2/3}}}, this means:


{{{rise/run=2/3}}}



which shows us that the rise is 2 and the run is 3. This means that to go from point to point, we can go up 2  and over 3




So starting at *[Tex \LARGE \left(0,-1\right)], go up 2 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+(2/2),2,2,90,270))
)}}}


and to the right 3 units to get to the next point *[Tex \LARGE \left(3,1\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)),
  blue(circle(3,1,.15,1.5)),
  blue(circle(3,1,.1,1.5)),
  blue(arc(0,-1+(2/2),2,2,90,270)),
  blue(arc((3/2),1,3,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=(2/3)x-1}}}


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



Now let's plot the two equations together:


{{{drawing(500,500,-2,22,-5,15,
  grid(1),
  graph(500,500,-2,22,-5,15,(1/2)x+2,(2/3)x-1)
)}}}

Graph of {{{y=(1/2)x+2}}} (red) and {{{y=(2/3)x-1}}} (green)



From the graph, we see that the two lines intersect at the point (18,11). So the solution is (18,11) which means that x=18 and y=11. 



So the system is consistent and independent.