Question 173156
Start with the given system of equations:



{{{system(3x-2y=6,3x+2y=6)}}}




Let's graph the first equation:



{{{3x-2y=6}}} Start with the first equation.



{{{-2y=6-3x}}} Subtract {{{3x}}} from both sides.



{{{-2y=-3x+6}}} Rearrange the terms.



{{{y=(-3x+6)/(-2)}}} Divide both sides by {{{-2}}} to isolate y.



{{{y=((-3)/(-2))x+(6)/(-2)}}} Break up the fraction.



{{{y=(3/2)x-3}}} Reduce.



Now let's graph the equation:





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


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



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




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



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


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



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



Now let's graph the second equation:



{{{3x+2y=6}}} Start with the second equation.



{{{2y=6-3x}}} Subtract {{{3x}}} from both sides.



{{{2y=-3x+6}}} Rearrange the terms.



{{{y=(-3x+6)/(2)}}} Divide both sides by {{{2}}} to isolate y.



{{{y=((-3)/(2))x+(6)/(2)}}} Break up the fraction.



{{{y=-(3/2)x+3}}} Reduce.





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


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



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




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



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


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-(3/2)x+3),
  blue(circle(0,3,.1)),
  blue(circle(0,3,.12)),
  blue(circle(0,3,.15)),
  blue(circle(2,0,.15,1.5)),
  blue(circle(2,0,.1,1.5)),
  blue(arc(0,3+(-3/2),2,-3,90,270)),
  blue(arc((2/2),0,2,2, 0,180))
)}}} So this is the graph of {{{y=-(3/2)x+3}}} through the points *[Tex \LARGE \left(0,3\right)] and *[Tex \LARGE \left(2,0\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,(3/2)x-3,-(3/2)x+3),
circle(2,0,0.08),
circle(2,0,0.10),
circle(2,0,0.12),
circle(2,0,0.14),
circle(2,0,0.16)
)}}} Graph of {{{y=(3/2)x-3}}} (red). Graph of {{{y=-(3/2)x+3}}} (green)



From the graph, we can see that the two lines intersect at the point *[Tex \LARGE \left(2,0\right)]. 



So the solution to the system of equations is *[Tex \LARGE \left(2,0\right)]. 



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