System of Equations: Solving for (x,y)

[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?

 

[Snazzy]

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

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

Start with that and let's see where it takes you.

 

[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.

 

[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 into x = 6-2y...
x = 6 - 2(12/5)
x = 1.2

So, (1.2, 2.4)

That seems right. Thank you.

 

[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.

 

[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 CHECKEverything is verified. Thanks.