Solving x & y from 2 Equations in Terms of a & b

[The legend]

Homework Statement

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

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

Homework Equations

the general algebra equations

The Attempt at a Solution

adding the equations

(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 don't know how to proceed..

 

[ehild]

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

ehild

 

[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 haven't got its solutions.. and I have never got such complex answers!)

 

[ehild]

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

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

 

[The legend]

Whew!
I tried it out your way and got the answer!(which I don't 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)

 

[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

 

[Mentallic]

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

 

[The legend]

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