Need Help With Solving Systems Using Substitution

I already have the answer, I just need someone to show me work to help me understand.

y=8-x
7=2-y

The answer should be (3, -5)

Just take 8 - x from the first equation and substitute it for y in the second one.

y = 8 - x
7 = 2 - y

7 = 2 - 8 + x
x = 7 - 2 + 8 = 13
y = 8 - x = - 5

since y=8-x you simply put a parenthesis where the y is in the other equation and but y=8-x. then you multiply 8 by 2 and -x by 2 and the equation should look like 7=16-2x then you move the 16 to the other side to get x by itself. 7-16=16-2x-16 so you should have -9=-2x. then you divide both sides by -2 and you have what x is. then put the eqaution x=whatever number you get into the top equation and solve the same way.

okay so all you have to do is solve for either x OR y in either of the two equations. This specific system is easy because the first equation is already solved for y, so less work needs to be done. Now plug in the value of y, which is (y = 8 - x) into the bottom equation.

This will give you: 7 = 2 - (8-x). Now you see that there is only one variable. Solve for the variable x...
7 = 2 - 8 + x
7 = -6 + x
13 = x (I see you made a typo because X CAN NOT equal 3)
Plug in the value of x (which is 13) into the first equation.... That will give you...
y = 8 - 13
y = -5