For example:

 

9y^2 - 6y - 9 - x = 0

 

What do you do in this case? Thanks!

Okay. I've always been good at figuring stuff out, but my training in math is quite limited. However, since nobody has responded yet...

 

Don't you just solve for one of the variables, then do a replacement? So, you say "X = blah blah blah" and then put "blah blah blah" in place of X and solve for Y. (note- X will equal something with a few Y's in it).

 

Make sense?

Then, once you know an actual number value for Y, you can put that into the equation and determine an actual number value for X.

The substitution method as you describe works with two-variable equations only if you're using a system of two or more equations; otherwise there are a large number of possible values for X and Y.

 

This leaves me curious why CrazCo even wants to complete the square in a case like this...

The question specifically required it. Thanks!

For example:

 

9y^2 - 6y - 9 - x = 0

 

What do you do in this case? Thanks!

 

Well, there isn't an x^2 term so it's not really possible to complete the square including the x, but you can complete the square with y...

 

First, rearrange the equation so that the three terms necessary to complete are on one side, and the x is on the other.

[math]9y^2 - 6y - 9 = x[/math]

Now, the closest square is [math](3y-1)^2 = 9y^2 - 6y + 1[/math]. This is 10 more than what you have on the left. So, if we add ten to both sides (preserving the equality) you get this:

[math]9y^2 - 6y - 9 + 10 = x + 10[/math]

[math]9y^2 - 6y + 1 = x + 10[/math]

But [math]9y^2 - 6y + 1 = (3y-1)^2[/math] so

[math](3y-1)^2 = x + 10[/math]

 

Now, this doesn't appear very useful if you're trying to solve for either variable, but if the directions were to just complete the square then I guess it works. Usually completing the square is a tool used to simplify an equation that isn't otherwise algebraically solvable.