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A friend of Mine gave me a system of two equations and asked me to solve them $\rightarrow$

$\sqrt{x}+y=11~~ ...1$ $\sqrt{y}+x=7~~ ...2$

I tried to solve them manually and got this horrendously complicated fourth degree equation $\rightarrow$

$\begin{align*} y &= (7-x)^2 ~...\mbox{(from 2)} \\ y &= 49 - 14 x + x^2 \\ \implies 11&= \sqrt{x}+ 49 - 14 x + x^2 ...(\mbox{from 1)}\\ \implies~~ 0&=x^4-28x^3+272x^2-1065x+1444 \end{align*}$

Solving this wasn't exactly my piece of cake but I could tell that one of Solutions would have been 9 and 4

But my friend kept asking for a formal solution.

I tried plotting the equations and here's what I got $\rightarrow$

enter image description here

So the equations had two pairs of solutions (real ones).

Maybe, Just maybe I think these could be solved using approximations.

So How do i solve them using a formal method (Calculus,Algebra,Real Analysis...)

P.S. I'm In high-school.

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    This is the same system as in http://math.stackexchange.com/questions/108962/2013-12-14

2 Answers 2

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Assume $x$ and $y$ are integers. Notice that, in this case, if $\sqrt x +y=11$, an integer, then $\sqrt x $ must be an integer. A similar argument can be made for $y$. So if they're integers then they're both perfect squares. Rephrasing in terms of the square roots (still integers) $X=\sqrt x,Y=\sqrt y$ $X+Y^2=11$ $Y+X^2=7$ subtracting the second equation from the first: $X-Y+Y^2-X^2=4$ $(X-Y)+(Y-X)(Y+X)=4$ $(Y-X)(X+Y-1)=4$ Both of the brackets are integers, so the only values they can take are the factors of $4$. So either $Y-X=2,X+Y-1=2$ or $Y-X=4,X+Y-1=1$ or$Y-X=1,X+Y-1=4$ Solving each of these is simple. The only one that gives positive integer values (the conditions of our little set up here) is the $3^{rd}$ one, which gives the answer you found. Keep in mind that there's nothing wrong with guessing and playing around with the problem first, then coming to a more structured argument later. If you want a full analytic solution you could use the quartic equation on the one you have and rule out the other solutions as involving the wrong branches of $\sqrt x$, but it's pointlessly messy.

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    right. $x$ and $y$ (which you found) are not negative. But plug them into the equations. They work for the first one, not for the second one. What does work for the second one and your numbers is $x-\sqrt y=7$. I.e. taking $-\sqrt y$. So you're correct in saying $y$ isn't negative. But it's the same thing as taking the negative square root of $4$ as $-2$. $4$ itself being positive doesn't change that.2012-06-01
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Once you guessed the solutions, you can easily prove that there are no others. Rewrite the equations as $y=11-\sqrt x=F(x)$ and $x=7-\sqrt y=G(y)$. Note that both $x,y\le 11$, so their square roots are at most $4$, which means that $x,y\ge 3$. Now just observe that $z\mapsto \sqrt z$ is a contraction on $[3,\infty)$ (the difference of values is less than the difference of arguments). Thus, $F$ and $G$ are also contractions whence if we had two different solutions $(x_1,y_1)$ and $(x_2,y_2)$, we would get $ |x_1-x_2|=|G(y_1)-G(y_2)|<|y_1-y_2|=|F(x_1)-F(x_2)|<|x_1-x_2| $ which is absurd.