Two algebraic problems I can't solve

1. Jul 24, 2012

stonecoldgen

1. The problem statement, all variables and given/known data
1) [(1/y2)-x2]/[(1/x2)-y2]=?

2) what is the sum of the solutions, x+y, in the following system of equations?
x/y=1/2
(x+2)/(y-9)=-4/5

3. The attempt at a solution

1) I started by separating the numerator of the initial big fractionk trying to simplify but failing. However, i did get to the point where I had to separated fractions with the same denominator, but it was a fail because I was just doing a round trip, lol.

2) I defined x in terms of y and plugged it in into the (x+2)/(y-9)=-4/5 equation, being as careful as I could with having all addingg/subtracting terms with the same denominator, careful about signs etc. But I'm just doing something wrong.

BTW, I'm already a college freshman, I already passed algebra back in 9th and 10th, but Im just loosing the rhythm for this stuff and I don't want that, so any help is appreciated, thanks (just tell me wich is the best path to start with if you can, thanks).

2. Jul 24, 2012

micromass

Staff Emeritus
It would be nice of you to actually type out what you did!

3. Jul 24, 2012

SammyS

Staff Emeritus
For #1, multiply the numerator & denominator by x2y2.

4. Jul 24, 2012

Mentallic

If you're in college, have you learnt Linear Algebra? You know, using matrices to solve linear equations? They're not necessary here, but it would be good practice to use them to solve #2 if you should already know them.

To start off, simplify both given equations into a linear combination of x and y, for example, $$\frac{x}{y}=\frac{1}{2}$$
$$2x=y$$
$$2x-y=0$$

Do the same for the second equation.

Now, we have a 3rd equation to solve, x+y=? but rather than a question mark, we'll introduce another variable, namely z for consistency.
You'll now have 3 linear equations with 3 unknowns. So either by using matrices or the conventional plugging and solving, try find the value of z.

5. Jul 25, 2012

stonecoldgen

hey this is pretty interesting! Im starting college today actually! I will see linear algebra next semester, but thanks a lot