System of equations problem

1. Apr 7, 2008

prophet05

[SOLVED] System of equations problem

I'm trying to brush up on my algebra and had difficulties with a problem.

Solve for (x,y): (x^-(2/3))(y^(2/3)) = (2x^(1/3))(y^-(1/3)) and 6 = x + 2y

I'm mainly having problems simplifying the first equation for y but got y=0. Any help?

2. Apr 7, 2008

Snazzy

$$x^{-a} = \frac{1}{x^a}$$

$$\frac{x^a}{x^b} = x^{a-b}$$

3. Apr 7, 2008

Dr Zoidburg

solving the second equation is easy enough - just make y the subject.

As for the first, I find it's easier to rewrite it without using negative powers.
$$x^{-2/3} = \frac{1}{x^{2/3}}$$

From there, it's just a matter of rearranging to make y the subject again.
From there, you can find x using both equations.

4. Apr 7, 2008

prophet05

Perfect, I think it was the basic concepts that helped. Just to verify:

I was able to simplify the first equation to y = 2x. Using that...
y = 2(6-2y)
y = 12 - 4y
5y= 12
y=12/5

Plug in to x = 6-2y...
x = 6 - 2(12/5)
x = 1.2

So, (1.2, 2.4)

That seems right. Thank you.

5. Apr 7, 2008

Snazzy

Plug in those values back into the original question to see if the left hand side equals the right hand side to make sure you're right.

6. Apr 7, 2008

prophet05

(1.2,2.4)

[y=2x]
2.4 = 2 (1.2)
2.4 = 2.4 CHECK

[x=6-2y]
1.2 = 6 - 2(2.4)
1.2 = 1.2 CHECK

Everything is verified. Thanks.