Question 199610
Start with the given system of equations:



{{{system(3x-y=-3,2x+y=-7)}}}



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



Let's graph the first equation:



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



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



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



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



Looking at {{{y=3x+3}}} we can see that the equation is in slope-intercept form {{{y=mx+b}}} where the slope is {{{m=3}}} 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,-10,10,
  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 {{{3}}}, this means:


{{{rise/run=3/1}}}



which shows us that the rise is 3 and the run is 1. This means that to go from point to point, we can go up 3  and over 1




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

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


and to the right 1 unit to get to the next point *[Tex \LARGE \left(1,6\right)]

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15)),
  blue(circle(1,6,.15,1.5)),
  blue(circle(1,6,.1,1.5)),
  blue(arc(0,3+(3/2),2,3,90,270)),
  blue(arc((1/2),6,1,2, 180,360))
)}}}



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


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,3x+3),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15)),
  blue(circle(1,6,.15,1.5)),
  blue(circle(1,6,.1,1.5)),
  blue(arc(0,3+(3/2),2,3,90,270)),
  blue(arc((1/2),6,1,2, 180,360))
)}}} So this is the graph of {{{y=3x+3}}} through the points *[Tex \LARGE \left(0,3\right)] and *[Tex \LARGE \left(1,6\right)]




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



Now let's graph the second equation:



{{{2x+y=-7}}} Start with the second equation.



{{{y=-7-2x}}} Subtract {{{2x}}} from both sides.



{{{y=-2x-7}}} Rearrange the terms and simplify.



Looking at {{{y=-2x-7}}} 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=-7}}} 



Since {{{b=-7}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-7\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,-7\right)]


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-7,.1)),
  blue(circle(0,-7,.12)),
  blue(circle(0,-7,.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 down 2  and over 1




So starting at *[Tex \LARGE \left(0,-7\right)], go down 2 units 

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


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

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-7,.1)),
  blue(circle(0,-7,.12)),
  blue(circle(0,-7,.15)),
  blue(circle(1,-9,.15,1.5)),
  blue(circle(1,-9,.1,1.5)),
  blue(arc(0,-7+(-2/2),2,-2,90,270)),
  blue(arc((1/2),-9,1,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-2x-7}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-2x-7),
  blue(circle(0,-7,.1)),
  blue(circle(0,-7,.12)),
  blue(circle(0,-7,.15)),
  blue(circle(1,-9,.15,1.5)),
  blue(circle(1,-9,.1,1.5)),
  blue(arc(0,-7+(-2/2),2,-2,90,270)),
  blue(arc((1/2),-9,1,2, 0,180))
)}}} So this is the graph of {{{y=-2x-7}}} through the points *[Tex \LARGE \left(0,-7\right)] and *[Tex \LARGE \left(1,-9\right)]



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



Now let's graph the two equations together:



{{{drawing(500,500,-10,10,-10,10,
grid(1),
graph(500,500,-10,10,-10,10,3x+3,-2x-7)
)}}} Graph of {{{y=3x+3}}} (red). Graph of {{{y=-2x-7}}} (green)



From the graph, we can see that the two lines intersect at the point *[Tex \LARGE \left(-2,-3\right)]. So the solution to the system of equations is *[Tex \LARGE \left(-2,-3\right)]. 


This means that the system has one unique solution.


This also tells us that the system of equations is consistent and independent.