# 2 variables!

1. Jul 13, 2010

### The legend

1. The problem statement, all variables and given/known data

Find x and y in terms of a and b from the following equations

x * √x + y * √y = a
x * √y + y * √x = b

2. Relevant equations
the general algebra equations

3. The attempt at a solution

(x + y) * (√x + √y) = a + b
x + y =
a + b
_______
√x + √y​

x =
a + b + y√x + y√y)
______________________
√x + √y​

Now substituting in eqn 1

a + b + y√x + y√y
_________________ + y√y = a
√x + √y

a + b + y√x + y√y + y√x√y + y2 = a√x + a√y

... Now I dunno how to proceed..

2. Jul 14, 2010

### ehild

To begin, add 3 times the second equation to the first one.

ehild

3. Jul 14, 2010

### The legend

I tried it out..

x * √x + y * √y = a
3*x * √y + 3*y * √x = b

adding and using (p + q)3 formula

(√x + √y)3 = 3b + a

therefore

√x = (3√3b + a) - √y
so
x√x = {3(√3b + a) - √y}3

Further solving

I get y =
-(3√3b + a)2 +- √(3√3b + a)4 + 4b * (3√3b + a)
_________________________________________________________________
2 * (3√3b + a)​

Is it right?(Because I havent got its solutions.. and I have never got such complex answers!!)

4. Jul 14, 2010

### ehild

Correct up to here, but all the others are wrong. What is x+y? The whole left side should be under the cubic root.

More: You can factor out √x √y from the second equation. As you know
(√x + √y), you get an equation for √x √y.

There will be a system of two equations, one for (√x + √y), the other for √x √y, which is easy to solve.

ehild

5. Jul 14, 2010

### The legend

Whew!
I tried it out your way and got the answer!(which I dont think I want to key down)

Though u say that the whole left side should be under cube root... actually it is(that is cube root of (√x + √y)3 so it will be just √x + √y)... so anything wrong? ( maybe I went wrong in the calculations)

6. Jul 14, 2010

### ehild

I mixed my hands... :) I meant the right side.

√x = (3√3b + a) - √y--this is wrong. You have the cubic root of 3 alone.

$$\sqrt{x}+\sqrt{y}=\sqrt[3]{3b+a}$$

ehild

7. Jul 14, 2010

### Mentallic

I think that's what he implied by the use of brackets (even though it wasn't correctly expressed)

8. Jul 15, 2010

### The legend

Ok thanks ehild!
And also Mentallic got me right! I juggled up my use of brackets!!