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System of equations problem

  • Thread starter prophet05
  • Start date
  • #1
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[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?
 

Answers and Replies

  • #2
458
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[tex]x^{-a} = \frac{1}{x^a}[/tex]

[tex]\frac{x^a}{x^b} = x^{a-b}[/tex]

Start with that and let's see where it takes you.
 
  • #3
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.
[tex]x^{-2/3} = \frac{1}{x^{2/3}}[/tex]

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
12
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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
458
0
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
12
0
(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.
 

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