Algebra 1 help (step by step please)

Hi I need some help w/ these algebra 1 problems. Can you give me the answers and show work/step by step way to solve them. Ok heres the problems-

1) Use substitution to solve the linear system.
3x-y=15
x+2y=(-2)

2) Solve the linear system by any method (substition/linear combination)
5x-2y=3
(-x)+6y=(-2)

3) Determine if the system has no solutions, one solution, or many solutions (show why)
3x+y=3
(-9x)-3y=(-9)

4) Find all solutions of the system: many, none, one (show why)
3x-y=2
(-3x)+y=(-2)

Thanks so much if you can help I really appreciate it!

In all 4 cases, let the top equation be EQ'A and
let the bottom equation be EQ'B

1) rearrange EQ'B, we get x = -2 - 2y
SUBSTITUTE the solve for x into EQ'A
3(-2 - 2y) - y = 15
-6 - 6y - y = 15
-7y = 21
y = -3

Since we know
x = -2 - 2y
x = -2 - 2(-3)
x = -2 + 6
x = 4

(4, -3)
---------------------------------
2) Linear Combined:
multiply by 3 to EQ'A, to get
15x - 6y = 9
(-x) + 6y = (-2)
Add EQ'A and EQ'B
14x = 7
x = 1/2
Substitute x = -1/2 into EQ'B
-(1/2) + 6y = -4/2
6y = -3/2
y = -3/12
y = -1/4

(1/2 , -1/4)
--------------------------------------…
3) Multiply EQ'A by - 3, we get
EQ'A" ={ -9x - 3y = -9 } which is equal to EQ'B
So they are the same equation
The answer is "Many solutions"
--------------------------------------…
4) If we add EQ'A and EQ'B together, we get 0
"Many solutions"