Question 145290
note: I added the solution to #2

<a name="one">


# 1    

<a href="#two">Jump to problem #2</a>

Once again, I'll do the first one to get you started (hopefully in the right direction)



Let's graph the first equation {{{-x + y = -1}}}. To do this, we must first solve for y



{{{-x+y=-1}}} Start with the first equation



{{{y=-1+x}}} Add {{{ x}}} to both sides



{{{y=x-1}}} Rearrange the equation






Looking at {{{y=x-1}}} 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=-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 {{{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,-1\right)], go up 1 unit 

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


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



Now draw a line through these points to graph {{{y=x-1}}}


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




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




Now let's graph the second equation {{{x + y = 3}}}. To do this, we must first solve for y





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



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



{{{y=-x+3}}} Rearrange the equation





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,-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 {{{-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(0,3\right)], go down 1 unit 

{{{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+(-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,-10,10,
  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, 0,180))
)}}}



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


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-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, 0,180))
)}}} 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,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,x-1,-x+3)
)}}} Graph of {{{y=x-1}}} (red) and {{{y=-x+3}}} (green)



So from the graph, we can see that the two lines intersect at the point (2,1).








<hr>



<a name="two">
# 2


<a href="#one">Jump to problem #1</a>


Let's graph the first equation {{{3x + y = -6}}}. To do this, we must first solve for y




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



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



{{{y=-3x-6}}} Rearrange the equation





Looking at {{{y=-3x-6}}} 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=-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 {{{-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 down 3  and over 1




So starting at *[Tex \LARGE \left(0,-6\right)], go down 3 units 

{{{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+(-3/2),2,-3,90,270))
)}}}


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,-6,.1)),
  blue(circle(0,-6,.12)),
  blue(circle(0,-6,.15)),
  blue(circle(1,-9,.15,1.5)),
  blue(circle(1,-9,.1,1.5)),
  blue(arc(0,-6+(-3/2),2,-3,90,270)),
  blue(arc((1/2),-9,1,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-3x-6}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-3x-6),
  blue(circle(0,-6,.1)),
  blue(circle(0,-6,.12)),
  blue(circle(0,-6,.15)),
  blue(circle(1,-9,.15,1.5)),
  blue(circle(1,-9,.1,1.5)),
  blue(arc(0,-6+(-3/2),2,-3,90,270)),
  blue(arc((1/2),-9,1,2, 0,180))
)}}} So this is the graph of {{{y=-3x-6}}} through the points *[Tex \LARGE \left(0,-6\right)] and *[Tex \LARGE \left(1,-9\right)]



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



Let's graph the second equation {{{x + y = -4 }}}. To do this, we must first solve for y



{{{x+y=-4}}} Start with the second equation



{{{y=-4-x}}}  Subtract {{{ x}}} from both sides



{{{y=-x-4}}} Rearrange the equation






Looking at {{{y=-x-4}}} 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=-4}}} 



Since {{{b=-4}}} this tells us that the y-intercept is *[Tex \LARGE \left(0,-4\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,-4\right)]


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

{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  blue(circle(0,-4,.1)),
  blue(circle(0,-4,.12)),
  blue(circle(0,-4,.15)),
  blue(arc(0,-4+(-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,-4,.1)),
  blue(circle(0,-4,.12)),
  blue(circle(0,-4,.15)),
  blue(circle(1,-5,.15,1.5)),
  blue(circle(1,-5,.1,1.5)),
  blue(arc(0,-4+(-1/2),2,-1,90,270)),
  blue(arc((1/2),-5,1,2, 0,180))
)}}}



Now draw a line through these points to graph {{{y=-x-4}}}


{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-x-4),
  blue(circle(0,-4,.1)),
  blue(circle(0,-4,.12)),
  blue(circle(0,-4,.15)),
  blue(circle(1,-5,.15,1.5)),
  blue(circle(1,-5,.1,1.5)),
  blue(arc(0,-4+(-1/2),2,-1,90,270)),
  blue(arc((1/2),-5,1,2, 0,180))
)}}} So this is the graph of {{{y=-x-4}}} through the points *[Tex \LARGE \left(0,-4\right)] and *[Tex \LARGE \left(1,-5\right)]




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



{{{drawing(500,500,-10,10,-10,10,
  grid(1),
  graph(500,500,-10,10,-10,10,-3x-6,-x-4)
)}}} Graph of {{{y=-3x-6}}} (red) and {{{y=-x-4}}} (green)



So from the graph, we can see that the two lines intersect at the point (-1,-3).