Question 728205
<pre>
2x+y = -8
5x-y = -6

We get some points on each.

To get some points on 

2x+y = -8

We arbitrarily choose x=0

  2x+y = -8
2(0)+y = -8
   0+y = -8
     y = -8

So one point is (0,-8)

We arbitrarily choose y=0

  2x+y = -8
  2x+0 = -8
    2x = -8
     x = -4

So one point is (-4,0)

We arbitrarily choose y=6

  2x+y = -8
  2x+6 = -8
    2x = -14
     x = -7

So one point is (-7,6)

Let's plot those three points:

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),
circle(-7,6,.2), circle(-4,0,.2),circle(0,-8,.2) )}}}{{{matrix(15,1,

NOW,"LET's",GET,A,STRAIGHT,EDGE,AND,DRAW,A,STRAIGHT,LINE,THROUGH,THOSE,THREE,"POINTS--->")}}}{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),
circle(-7,6,.2), circle(-4,0,.2),circle(0,-8,.2),green(line(-16,24,21,-50)) )}}}

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

To get some points on 

5x-y = -6

We arbitrarily choose x=0

  5x-y = -6
5(0)-y = -6
   0-y = -6
    -y = -6
     y = 6

So one point is (0,6)

We arbitrarily choose y=1

  5x-y = -6
  5x-1 = -6
    5x = -5
     x = -1

So one point is (-1,1)

We arbitrarily choose x=-3

   5x-y = -6
5(-3)-y = -6
  -15-y = -6
     -y =  9
      y = -9

So one point is (-3,-9)

Let's plot those three points:

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),
circle(-7,6,.2), circle(-4,0,.2),circle(0,-8,.2),green(line(-16,24,21,-50)),
circle(-3,-9,.2), circle(-1,1,.2),circle(0,6,.2)


 )}}}{{{matrix(15,1,

NOW,"LET's",GET,A,STRAIGHT,EDGE,AND,DRAW,A,STRAIGHT,LINE,THROUGH,THOSE,THREE,"POINTS--->")}}}{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),
circle(-7,6,.2), circle(-4,0,.2),circle(0,-8,.2),green(line(-16,24,21,-50)),
red(line(-20,-94,19,101)),circle(-3,-9,.2), circle(-1,1,.2),circle(0,6,.2)
 )}}}

Now from the point where they cross, we draw one line straight to the
x-axis, and another line straight to the y-axis:

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),
circle(-7,6,.2), circle(-4,0,.2),circle(0,-8,.2),green(line(-16,24,21,-50)),
red(line(-20,-94,19,101)),circle(-3,-9,.2), circle(-1,1,.2),circle(0,6,.2),
blue(line(-2,-4,-2,0),line(-2,-4,0,-4))


 )}}}

We see that one blue line ends up at -2 on the x-axis, and the other
blue line ends up at -4 on the y-axis, which means the solution is
x=-2 and y=-4.  The point where the green and red line intersect is
the point (-2,-4).

Now we check to see if when we substitute -2 for x and -4 for y in
each we come out with a true equation.

Substituting them in

      2x+y = -8
2(-2)+(-4) = -8
      -4-4 = -8
        -8 = -8

That checks.

Substituting them in

      5x-y = -6
5(-2)-(-4) = -6
     -10+4 = -6
        -6 = -6

That also checks.  So we know that the solution is

(x,y) = (-2,-4)

Edwin</pre>