<s>[INST] Solving floor function system of equations given $x \lfloor y \rfloor + y \lfloor x \rfloor =66$ and $x \lfloor x \rfloor + y \lfloor y \rfloor=144$ 
$x$ and $y$ are real numbers satisfying $x \lfloor y \rfloor + y \lfloor x \rfloor =66$ and $x \lfloor x \rfloor  + y \lfloor y \rfloor=144.$ Find $x$ and $y.$

I first assumed $x>y.$ Then I approximated $x$ and $y$ respectively to be in ranges $11<x<12$ and $2<y<3.$ Then this means $\lfloor x \rfloor =11$ and $\lfloor y \rfloor =2.$ So I wrote the system of equations as $2x+11y=66$ and $11x+2y=144,$ but solving the system gave me solutions of $x$ and $y$ that didn't satisfy the $\lfloor x \rfloor =11$ and $\lfloor y \rfloor =2$ constraint.  I'm not really sure how to approach this problem from here other than trying values. May I have some help?
 [/INST] Here is an idea: you already showed that $(x+y)(\lfloor x \rfloor + \lfloor y \rfloor)=210$. If you denote $k=\lfloor x \rfloor + \lfloor y \rfloor$ then $k\leq x+y<k+2$. So we have $k^2\leq k(x+y)<k(k+2)$. Since $k(x+y)=210$, we now know that $k^2\leq 210<k(k+2)$. Solving this inequality for integer $k$ yields $k=14$.
Now, use the substruction part: $(x-y)(\lfloor x \rfloor - \lfloor y \rfloor)=78$. Denote $m=\lfloor x \rfloor - \lfloor y \rfloor$ and assume WLOG that $x \geq y$. Now $m-1<x-y<m+1$, so, similarly we will get that $m(m-1)<78<m(m+1)$. Solving this for integer $m$ yields $m=9$.
This seems impossible, as $\lfloor x \rfloor$ and $\lfloor y \rfloor$ are integers, so $m,k$ should have the same parity - so I guess there is a mistake somewhere, but the idea seems to have potential - that's why I decided to write it up.
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