A little help with math work please!

2x+3y=6
x-2y=10

Part 1: Solve the system and locate the intersection point. Type the ordered pair solution.
Part 2: Solve the system using the substitution method. Show all work here and indicate the solution for the system as an ordered pair.
Part 3: Solve the system using the addition method. Show all work here and indicate the solution for the system as an ordered pair

10pts thanks so much to anyone who can help. even if its only one question.

I'll do Parts 2 and 3 first.
Part 2: write the second equation as x = 10 + 2y
Substitute it in the first equation:
2 (10+2y) + 3y = 6
Simplify: 20+4y+3y = 6
20+7y = 6
so 7y =-14 and y = -2
then x = 10+2y = 10 -4 = 6
The ordered pair: (-2,6)

Part 3: Addition method: multiply second equation by 2 and subtract it from the first:
You get 7y = -14
The rest is as before.

The ordered pair is also the intersection point asked for in Part 1.

In Part 3, we subtracted, and you might ask "why is that the addition method?" Well, there's no such thing as a subtraction method. Subtraction is merely the same as adding the negative of something.

(1)
Using any method?
2x + 3y = 6 <=> x = (6 - 3y)/2
x - 2y = 10 <=> x = 10 + 2y
=>
(6 - 3y)/2 = 10 + 2y
6 - 3y = 20 + 4y
-14 = 7y
-2 = y
x = 10 + 2(-2) = 10 - 4 = 6
So the solution is (6,-2).

(2)
x = 10 + 2y
2(10 + 2y) + 3y = 6
20 + 4y + 3y = 6
7y = -14
y = -2
x = 10 + 2(-2) = 10 - 4 = 6
Solution: (6,-2).

(3)
2x + 3y = 6
x - 2y = 10
=>
2x + 3y = 6
-2x + 4y = -20
=>
7y = -14
y = -2
===
2x + 3y = 6
x - 2y = 10
=>
4x + 6y = 12
3x - 6y = 30
=>
7x = 42
x = 6
Solution: (6,-2)