# Solve this two equations with two unkown of different degrees

1. Nov 30, 2007

### racer

Hello there

Anyone who can solve this problem is to me a genius in math.

X^2 + Y^2 = 25
X^3 + Y^3 = 91

Find X and Y
I know that X or Y will be 3 or 4 but I don't know how to solve that equation.

thanks

2. Nov 30, 2007

### Xevarion

Let $$x+y = A$$ and $$xy = B$$. Then
$$A^2 - 2B = 25$$
$$A^3 - 3AB = 91$$
$$AB = 25A - 91$$
$$B = 25 - \frac{91}{A}$$
$$A^2 + \frac{192}{A} - 50 = 25$$
$$A^3 - 75A + 192 = 0$$
You already observed that $A=7$ is a solution, which helps us factor this.
$$(A - 7)(A^2 + 7A - 26) = 0$$
Now we can use the quadratic formula to do the last bit. The possibilities are
$$A = 7, \frac{-7 \pm 3\sqrt{17}}{2}$$

I don't really want to find the values of $x$ and $y$ that give the other two roots. But if you want to do it, just plug in to find $B$, and then you have values for X + Y and XY, so you can solve for X in terms of Y, plug in to the other equation, and solve for Y (it will be a quadratic).

Doing it for 7 would be like:
A = 7
So B = 25 - 91/7 = 25 - 13 = 12.
We have X + Y = 7 and XY = 12. Then X = 7 - Y, so (7-Y)Y = 12, or Y^2 - 7Y + 12 = 0, or (Y-3)(Y-4) = 0. Then Y = 3 or Y = 4, and that gives X = 4 or X = 3.

There may be a simpler way than what I did, but basically you're going to have to do a lot of algebraic manipulation no matter what.

3. Nov 30, 2007

### Feldoh

$$x^2 + y^2 = 25$$
$$y = \sqrt{25-x^2}$$ and $$y = -\sqrt{25-x^2}$$

Can you go from there?

4. Dec 1, 2007

### racer

Xevarion

Feldoh

I don't think so, I tried but usually it ends up 25 = 25 :D

5. Dec 1, 2007

### Feldoh

Xevarion's way is better however mine still works. You seem to be using the solutions of y I found to solve

$$y^2+x^2=25$$

Which will give you 25=25

6. Dec 1, 2007

### racer

Feldoh

Yeah, I've used the solutions of y you found.

7. Dec 1, 2007

### Xevarion

You can put it into the other equation. However I didn't do this way because then you get a very high degree equation with X. Even knowing the solutions X=3 and X=4 might leave it pretty hard to factor. But Feldoh's method probably can still work. (I'm just too lazy to try it :P)

The reason I introduced A and B was so that you can solve for things twice, which means each time the degree is smaller (3 for A and 2 for X, if you recall).

8. Dec 1, 2007

### Feldoh

Sorry to be offtopic but Xevarion do you post on the AOPS forums?

9. Dec 2, 2007

### racer

Xevarion

you wrote 192 instead of 182 but you apparently
calculated it using 182 because you wouldn't get that answer if did 192.

Your answer really taught me something, it is that math is not superficial science where someone just replace uknowns with their values but it is knowing things from every side and angle.

I am very impressed, do you have a B.S in math?

I think it works but the answer would be so long.

10. Dec 2, 2007

### arildno

One might wonder how Xevarion pulled that one out of his hat.
I suspect he essentially thought of the polynomial factorization:
$$x^{3}+y^{3}=(x+y)(x^{2}-xy+y^{2})$$
Thus, using x+y and xy as new variables is a quite natural choice.

Perhaps he had a different first idea..

11. Dec 2, 2007

### Xevarion

It's essentially the same as what I did. The factorizations of $$x^n + y^n$$ and $$x^n - y^n$$ are very useful.

racer: hehe, thanks. Actually my university gives an AB degree for math (what's called BA elsewhere maybe). And I'll have it in a year and a half, if all goes well. ;-)