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$\sqrt{x} +y = 4$, $\sqrt{y} +x= 6$, find the solution (x,y). $NOTE$ : $\sqrt{4}+1= 4-1$, $\sqrt{1} +4 =1+4$

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$(\sqrt{2}+1) \lt (2+\sqrt{1})$, $(\sqrt{4}+1) \lt (4+\sqrt{1})$ –  Rajesh K Singh Aug 9 '12 at 5:09
    
$\sqrt x=4-y$, $x=y^2-8y+16=6-\sqrt y$, $y=(y^2-8y+10)^2$ gives you an equation of degree 4 in $y$. If you can solve it, you win. –  Gerry Myerson Aug 9 '12 at 6:00
    
Another approach is $y=4-\sqrt x$, $\sqrt y = 6-x$ implies $y=(6-x)^2$. Since $y=y$, we can set the right side of both equations equal to each other. Then we get an 4th degree equation in $\sqrt x$. –  Matt Groff Aug 9 '12 at 6:07
    
let $u=\sqrt{x}, v=\sqrt{y}$. Then $u^2+v=4, u+v^2=6$ Hence $v=4-u^2$. Hence $u+16-8u^2+u^4=6 \iff u^4-8u^2+u+10=0$. Wolframalpha gives some disgusting solutions to this quartic equation: wolframalpha.com/input/?i=u%5E4-8u%5E2%2Bu%2B10%3D0 –  progressiveforest Aug 9 '12 at 6:07
    
Beautiful question with ugly answer... –  ᴊ ᴀ s ᴏ ɴ Aug 9 '12 at 6:13

2 Answers 2

This is basically the method which was suggested in the comments above - turning this into a quartic equation. We will see whether someone suggest a substantially more elegant solution.

$$\sqrt{x}+y=4\\ x+\sqrt{y}=6$$

Using the substitution $\sqrt{x}=s$ and $\sqrt{y}=t$ we get: $$s+t^2=4\\ t+s^2=6$$

Which gives $$s=4-t^2=4-(6-s^2)^2\\ (s^2-6)^2+s-4=0\\ s^4-12s^2+s+32=0$$

It should be possible to solve this as a quartic equation, although it would be quite laborious. You can check what WolframAlpha is able to find out here and here

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great attempt indeed –  Rajesh K Singh Aug 9 '12 at 6:20

Let, $\sqrt{x}=s$, $\sqrt{y}=t$

we have, $s^4 -12s^2+s+32=0$, which is a 'biquadratic' equation of the form,

$$(s^2+ks+l)(s^2-ks+m)=0$$

i.e. $$s^4 -12s^2+s+32=(s^2+ks+l)(s^2-ks+m)$$

now by equating coefficients, we have

$$l+m-k^2 = -12, k(m-l) = 1, lm = 32$$

from the first two of these equations, we obtain

$$2m=k^2-12+(1/k), 2l=k^2-12-(1/k)$$

hence substituting in the third equation, the values of l,m,

$$(4)(32)=(k^2-12-1/k)(k^2-12+1/k)$$

$$128=(k^2-12)^2-1/k^2$$

$$128=k^4+144-24k^2-1/k^2$$

$$k^6-24k^4+16k^2-1=0$$

this is a cubic in $k^2$ which always has one real positive solution and we can find $k^{2}$,$l$,$m$

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