$\sqrt{x} +y = 4$, $\sqrt{y} +x= 6$, 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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0graphical plot will give quick and approximate solution – 2012-08-09
2 Answers
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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0great attempt indeed – 2012-08-09
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 $k^{2}$,$l$,$m$