Help rearranging this equation

Can someone help me understand how to rearrange this equation to get the following answer(s)

(x^p_y^p)^((1/p)-1) * x^(p-1) - Lambda*price1 = 0

(x^p_y^p)^((1/p)-1) * y^(p-1) - Lambda*price2 = 0

p*x + p*x = y = 0

which are my Lagrange multipliers for the cobb douglas function (x^p+y^p)^(1/p) where p is alpha/beta

the answer is

**x=y(price1/price2)^(1/(p-1)**

lambda = price1x + price2y

My basic math is not up the scratch so could somebody help me understand how to solve for x in the equations given

thanks

Re: Help rearranging this equation

I don't think I understand your entire question, but you have:

$\displaystyle (x^py^p)^{\frac{1}{p}-1}x^{p-1} - \lambda{price_1} = 0$

$\displaystyle (x^py^p)^{\frac{1}{p}-1}y^{p-1} - \lambda{price_2} = 0$

Add $\displaystyle \lambda{price_1}$ to both sides of the first equation and $\displaystyle \lambda{price_2}$ to both sides of the second, resulting in:

$\displaystyle (x^py^p)^{\frac{1}{p}-1}x^{p-1} = \lambda{price_1}$

$\displaystyle (x^py^p)^{\frac{1}{p}-1}y^{p-1} = \lambda{price_2}$

And then divide the first by the second, cancelling $\displaystyle \lambda$ and $\displaystyle (x^py^p)^{\frac{1}{p}-1}$, giving:

$\displaystyle \left(\frac{x}{y}\right)^{p-1}=\frac{price_1}{price_2}$

and then solve for x.

- Hollywood