Solve systems by substitution
x+y=-1
3x-2y=2
and
x=y-2
3x-4y=-13

Problem one: 
1.) Isolate one variable of one equation
x+y-x=-1-x
y=-1-x
2.) Plug in equation for isolated value into other original equation
3x-2(-1-x)=2
3.) Distribute
3x+2+2x=2
4.) Add like terms
5x+2=2
5.) Subtract two from both side
5x=0
6.) Divide both sides by 5
x=0
7.) Plug in numerical value for x into one equation (original or not)
0+y=1
8.) Subtract zero from both sides
y=1
For the first problem, the answer is: x=0, y = 1

For the second problem, follow very similar steps:
3(y-2)-4y=-13
3y-6-4y=-13
-y-6=-13
-y=-7
y=7

x=y-2
x=7-2
x=5
That's your answer for the second problem: x=5, y=7

Hope this helps!

1x + 1y = -1 ⇒ 3x - 3y = -3
3x -  2y = 2  ⇒ 3x - 2y = 2
                                -y = -5
                                  y = 5
                       3x - 2(5) = 2
                         3x - 10 = 2
                              +10  +10
                                3x = 12
                                 3      3
                                 x = 4
                          (x , y) = (4, 5)

1x - 1y = -2   ⇒ 3x - 3y = -6
3x - 4y = -13 ⇒ 3x - 4y = -13
                                   y = 7
                        3x - 4(7) = -13
                         3x - 28 = -13
                              +28    +28
                                3x = 15
                                 3      3
                                  x = 5
                            (x, y) = (5, 7)

Um what? rephrase that ?