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$$\sqrt{x}+y=4\tag{A}$$

$$x+\sqrt{y}=6\tag{B}$$

Subtracting A from B, we have

$$x-y -\sqrt{x}+\sqrt{y}=2$$

$$(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}-1)=2$$

$$(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}-1)=(\sqrt{2})(\sqrt{2})$$

Now we have,

$$\sqrt{x}-\sqrt{y}=\sqrt{2}\tag{AA}$$

$$\sqrt{x}+\sqrt{y}-1=\sqrt{2}\tag{BB}$$

This gives,

$$\sqrt{x}=\sqrt{2}+\frac{1}{2}$$

$$\sqrt{y}=\frac{1}{2}$$

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  • $\begingroup$ Note that since we don't have any restriction we also could have written the factors as: $\sqrt{4} \cdot 1$ Similarly we could write it in many forms. $\endgroup$
    – Adienl
    Jul 16, 2013 at 14:21
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    $\begingroup$ $(AA)$ and $(BB)$ together would imply that $\sqrt{x}=\frac12\sqrt{2}+\frac12$. But another problem is that if you multiply two numbers and get $2$, that doesn't necessarily mean that each of the two numbers you multiplied is $\sqrt{2}$. Also, you should try checking your solution by substituting it into the two equations you started with. $\endgroup$ Jul 16, 2013 at 14:28

3 Answers 3

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Do a substitution $x= X^2$ and $y=Y^2$, to get 2 quadratic equations. We have the conditions that $x , y \geq 0$, which also imply that $y\leq 4, x \leq 6$.

Substitute one into the other and you a get $ (6-x^2)^2 = 4-x$.

Solve this quartic (which doesn't seem to have nice roots), subject to $ 0 \leq x \leq 4$. You will get 2 solutions. Verify if they work.

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  • $\begingroup$ That's the approach I took, too. There seems to be no "easy" way around solving the quartic in the valid domain! $\endgroup$
    – amWhy
    Jul 16, 2013 at 14:39
  • $\begingroup$ yeah, i get stuck at that point too. I'd just plug it into wolfram alpha $\endgroup$
    – Zar
    Jul 16, 2013 at 14:39
  • $\begingroup$ nice approach indeed !! $\endgroup$ Jul 16, 2013 at 14:45
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That answer doesn't work, which you can see if you plug your values into your original equations. You made the mistake when you said

$\sqrt{x}-\sqrt{y}=\sqrt{2}$ -- AA

$\sqrt{x}+\sqrt{y}-1=\sqrt{2}$ -- BB

this doesn't follow from the previous equation because they don't both have to equal the squareroot of 2. They just have to multiply to 2. You are trying to solve 2 variable with one equation at this point. You lose information from the previous equations when you do this, so your final answer didn't work. What you might want to try instead is a substitution method, although this ends up with some very messy equations.

edit: so far i end up with

$0 = y^2 - 8y + \sqrt y + 10$

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  • $\begingroup$ sorry if i didn't help you completely answer it. I'm still working on solving it. $\endgroup$
    – Zar
    Jul 16, 2013 at 14:30
  • $\begingroup$ wonderful effort. $\endgroup$ Jul 16, 2013 at 14:42
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The solution as given by wolfram is as follows,

enter image description here

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  • $\begingroup$ unfortunately this doesn't get all the answers, because wolfram alpha only looks at half the graph for squareroots. There are 3 more for the negative squareroots $\endgroup$
    – Zar
    Jul 16, 2013 at 15:39

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