$\sqrt{x} +y = 4$, $\sqrt{y} +x= 6$, find the solution (x,y). $NOTE$ : $\sqrt{4}+1= 4-1$, $\sqrt{1} +4 =1+4$
$\sqrt{x} +y = 4$, $x+ \sqrt{y}= 6$, find the solution $(x,y)$
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algebra-precalculus
systems-of-equations
radicals
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0$(\sqrt{2}+1) \lt (2+\sqrt{1})$, $(\sqrt{4}+1) \lt (4+\sqrt{1})$ – 2012-08-09
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0$\sqrt x=4-y$, $x=y^2-8y+16=6-\sqrt y$, $y=(y^2-8y+10)^2$ gives you an equation of degree 4 in $y$. If you can solve it, you win. – 2012-08-09
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0Another 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 $\sqrt x$. – 2012-08-09
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0let $u=\sqrt{x}, v=\sqrt{y}$. 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: http://www.wolframalpha.com/input/?i=u%5E4-8u%5E2%2Bu%2B10%3D0 – 2012-08-09
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1Beautiful question with ugly answer... – 2012-08-09
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0graphical plot will give quick and approximate solution – 2012-08-09