you would think...?

You have two options for solving this--(1) substitution, and (2) combination. The first one is quicker to explain and easier to type out, so that's what I'm suggesting.

The deal with substitution is that you have two variables (x and y) and two equations with each of them in them. So you can solve this system!

First, solve one of the equations for one of the variables--that is, get the one variable on one side, and everything else on the other side. The second equation looks easiest to me:

22y = x + 25

(either subtract 25 from both sides to get x alone, or divide both sides by y to get y alone--it'll work either way!)

22y - 25 = x

Now plug in what x is equal to in the other equation.

-9x - 5y = 22

-9(22y - 25) - 5y = 22

-198y + 225 - 5y = 22

-203y = 203

y = 1

Finally, plug in the value for y into the FIRST equation (the one you first solved for x) and then solve for an actual value of x.

y = 1

x = 22y - 25

x = 22(1) - 25

x = 22-25

x = -3

So now you have your solution, x = -3, y = 1, which you write as the coordinate pair, (-3, 1).

-9x-5y=22

22y = x + 25

22y =x + 25

22y- 25 = x

-9 (22y-25) - 5y = 22

-198 y + 225 -5y = 22

-203y + 225 =22

-203y = -203

y= 1

-9x -5y = 22

-9x -5 (1) = 22

-9x -5 =22

-9x =27

x= -3

y = 1

x = -3