Simple Simultaneous equation help

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The discussion focuses on solving a system of simultaneous equations involving variables \(x_1\), \(x_2\), and a parameter \(\lambda\). The key equations presented are \(w_1 - \lambda x_2 = 0\), \(w_2 - \lambda x_1 = 0\), and \(\bar y - x_1x_2 = 0\). The correct solution for \(\lambda^*\) is established as \(\lambda^* = \sqrt{\frac{w_1w_2}{\bar y}}\). The final solution for \(x^*\) is confirmed to be \(x^* = \left(\sqrt{\frac{w_2 \bar{y}}{w_1}}, \sqrt{\frac{w_1 \bar{y}}{w_2}}\right)\).

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nacho-man
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Assuming $x_1, x_2 \geq 0, \lambda \neq 0, w_1,w_2 > 0$

We have the equalities:

$$w_1 - \lambda x_2 = 0 ... (1)$$
$$w_2 - \lambda x_1 = 0 ... (2)$$
$$\bar y - x_1x_2= 0 ... (3)$$

My solutions say that $\lambda^* = \sqrt\frac{w_1w_2}{\bar y}$
Which I was able to solve myself.

The other solution is $x^* = (\frac{w_2 \bar y}{w_1}, \frac{w_1 \bar y}{w_2})$

Which I cannot seem to get. Would anyone be so kind as to point out how to obtain solutions for $x^*$ ?

I tried:

$x_1= \frac{w_2}{\lambda}$ and $x_2 = \frac{w_1}{\lambda}$

then put these into the 3rd equation, but ended up gettig
$\bar y = \frac{w_1w_2}{\lambda^2}$ which I couldn't see turning into what I needed either way - even just subbing in one of the x-values at a time, but to no avail.
 
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anyone? :(

If I wasn't clear, $x^* = (x_1,x_2)$
 
Hi nacho,

The square root signs are missing in the answer for $x^*$. It should be $x^* = (\sqrt{w_2 \bar{y}/w_1}, \sqrt{w_1\bar{y}/w_2})$.
 
this was a suspicion of mine - thank you!
 

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