Question 199578
Start with the given system of equations:



{{{system(x+y=15,x-y=3)}}}



In order to graph these equations, we <font size="4"><b>must</b></font> solve for y first.



Let's graph the first equation:



{{{x+y=15}}} Start with the first equation.



{{{y=15-x}}} Subtract {{{x}}} from both sides.



{{{y=-x+15}}} Rearrange the terms and simplify.




Looking at {{{y=-x+15}}} 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=15}}} 



Since {{{b=15}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,15\right)].Remember the y-intercept is the point where the graph intersects with the y-axis


So we have one point *[Tex \LARGE \left(0,15\right)]


{{{drawing(500,500,-10,10,-4,16,
  grid(1),
  blue(circle(0,15,.1)),
  blue(circle(0,15,.12)),
  blue(circle(0,15,.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 down 1  and over 1




So starting at *[Tex \LARGE \left(0,15\right)], go down 1 unit 

{{{drawing(500,500,-10,10,-4,16,
  grid(1),
  blue(circle(0,15,.1)),
  blue(circle(0,15,.12)),
  blue(circle(0,15,.15)),
  blue(arc(0,15+(-1/2),2,-1,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,14\right)]

{{{drawing(500,500,-10,10,-4,16,
  grid(1),
  blue(circle(0,15,.1)),
  blue(circle(0,15,.12)),
  blue(circle(0,15,.15)),
  blue(circle(1,14,.15,1.5)),
  blue(circle(1,14,.1,1.5)),
  blue(arc(0,15+(-1/2),2,-1,90,270)),
  blue(arc((1/2),14,1,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-x+15}}}


{{{drawing(500,500,-10,10,-4,16,
  grid(1),
  graph(500,500,-10,10,-4,16,-x+15),
  blue(circle(0,15,.1)),
  blue(circle(0,15,.12)),
  blue(circle(0,15,.15)),
  blue(circle(1,14,.15,1.5)),
  blue(circle(1,14,.1,1.5)),
  blue(arc(0,15+(-1/2),2,-1,90,270)),
  blue(arc((1/2),14,1,2, 0,180))
)}}} So this is the graph of {{{y=-x+15}}} through the points *[Tex \LARGE \left(0,15\right)] and *[Tex \LARGE \left(1,14\right)]



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



Now let's graph the second equation:



{{{x-y=3}}} Start with the second equation.



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



{{{y=(3-x)/(-1)}}} Divide both sides by {{{-1}}} to isolate {{{y}}}.



{{{y=x-3}}} Rearrange the terms and simplify.




Looking at {{{y=x-3}}} 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=-3}}} 



Since {{{b=-3}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-3\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,-3\right)]


{{{drawing(500,500,-10,10,-4,16,
  grid(1),
  blue(circle(0,-3,.1)),
  blue(circle(0,-3,.12)),
  blue(circle(0,-3,.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,-3\right)], go up 1 unit 

{{{drawing(500,500,-10,10,-4,16,
  grid(1),
  blue(circle(0,-3,.1)),
  blue(circle(0,-3,.12)),
  blue(circle(0,-3,.15)),
  blue(arc(0,-3+(1/2),2,1,90,270))
)}}}


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,-2\right)]

{{{drawing(500,500,-10,10,-4,16,
  grid(1),
  blue(circle(0,-3,.1)),
  blue(circle(0,-3,.12)),
  blue(circle(0,-3,.15)),
  blue(circle(1,-2,.15,1.5)),
  blue(circle(1,-2,.1,1.5)),
  blue(arc(0,-3+(1/2),2,1,90,270)),
  blue(arc((1/2),-2,1,2, 180,360))
)}}}



Now draw a line through these points to graph {{{y=x-3}}}


{{{drawing(500,500,-10,10,-4,16,
  grid(1),
  graph(500,500,-10,10,-4,16,x-3),
  blue(circle(0,-3,.1)),
  blue(circle(0,-3,.12)),
  blue(circle(0,-3,.15)),
  blue(circle(1,-2,.15,1.5)),
  blue(circle(1,-2,.1,1.5)),
  blue(arc(0,-3+(1/2),2,1,90,270)),
  blue(arc((1/2),-2,1,2, 180,360))
)}}} So this is the graph of {{{y=x-3}}} through the points *[Tex \LARGE \left(0,-3\right)] and *[Tex \LARGE \left(1,-2\right)]



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



Now let's graph the two equations together:



{{{drawing(500,500,-10,10,-4,16,
grid(1),
graph(500,500,-10,10,-4,16,-x+15,x-3)
)}}} Graph of {{{y=-x+15}}} (red). Graph of {{{y=x-3}}} (green)



From the graph, we can see that the two lines intersect at the point *[Tex \LARGE \left(9,6\right)]. So the solution to the system of equations is *[Tex \LARGE \left(9,6\right)]. This tells us that the system of equations is consistent and independent.