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?
 A: By adding the given equations,
$$\begin{align*}
(x+y)(\lfloor x\rfloor + \lfloor y\rfloor) &= 210\\
(\lfloor x\rfloor+\lfloor y\rfloor + \{x\}+\{y\})(\lfloor x\rfloor +\lfloor y\rfloor) &= 210\\
\left(\lfloor x\rfloor+\lfloor y\rfloor+\frac{\{x\}+\{y\}}2\right)^2 - \left(\frac{\{x\}+\{y\}}2\right)^2 &= 210\\
\end{align*}$$
Since $0\le \{\ldots\} < 1$, then $0\le \frac{\{x\}+\{y\}}2 < 1$ and $0\le \left(\frac{\{x\}+\{y\}}2\right)^2 < 1$,
$$
210 \le \left(\lfloor x\rfloor+\lfloor y\rfloor+\frac{\{x\}+\{y\}}2\right)^2 < 211\\
\begin{align*}
14 < \lfloor x\rfloor+\lfloor y\rfloor+\frac{\{x\}+\{y\}}2 &< 15 &\text{or}&&-15 < \lfloor x\rfloor+\lfloor y\rfloor+\frac{\{x\}+\{y\}}2 &< -14\\
13 < \lfloor x\rfloor+\lfloor y\rfloor &< 15 &\text{or}&&-16 < \lfloor x\rfloor+\lfloor y\rfloor &< -14\\
\lfloor x\rfloor+\lfloor y\rfloor &=14 &\text{or}&& \lfloor x\rfloor+\lfloor y\rfloor &= -15\\
\end{align*}$$
Similarly, by subtracting the given equations,
$$\begin{align*}
(x-y)(\lfloor x\rfloor - \lfloor y\rfloor) &= 78\\
(\lfloor x\rfloor-\lfloor y\rfloor + \{x\}-\{y\})(\lfloor x\rfloor -\lfloor y\rfloor) &= 78\\
\left(\lfloor x\rfloor-\lfloor y\rfloor+\frac{\{x\}-\{y\}}2\right)^2 - \left(\frac{\{x\}-\{y\}}2\right)^2 &= 78\\
\end{align*}$$
Since $0\le \{\ldots\} < 1$, then $-\frac12 < \frac{\{x\}-\{y\}}2 < \frac12$ and $0\le \left(\frac{\{x\}-\{y\}}2\right)^2 < \frac14$,
$$
78 \le \left(\lfloor x\rfloor-\lfloor y\rfloor+\frac{\{x\}-\{y\}}2\right)^2 < 78.25\\
\begin{align*}
8.8 < \lfloor x\rfloor-\lfloor y\rfloor+\frac{\{x\}-\{y\}}2 &< 9 &\text{or}&&-9 < \lfloor x\rfloor-\lfloor y\rfloor+\frac{\{x\}-\{y\}}2 &< -8.8\\
8.3 < \lfloor x\rfloor-\lfloor y\rfloor &< 9.5 &\text{or}&&-9.5 < \lfloor x\rfloor-\lfloor y\rfloor &< -8.3\\
\lfloor x\rfloor-\lfloor y\rfloor &=9 &\text{or}&& \lfloor x\rfloor-\lfloor y\rfloor &= -9\\
\end{align*}$$
Considering the parity of $\lfloor x\rfloor$ and $\lfloor y \rfloor$, since their difference ($=9$) is odd, their sum is also odd:
$$\begin{align*}
\lfloor x\rfloor+\lfloor y\rfloor &= -15\\
\lfloor x\rfloor-\lfloor y\rfloor &= \pm 9\\
(\lfloor x\rfloor, \lfloor y \rfloor) &= (-3, -12) &\text{or}&& (\lfloor x\rfloor, \lfloor y \rfloor) &= (-12, -3)
\end{align*}$$
Substituting the first case $(\lfloor x\rfloor, \lfloor y \rfloor) = (-3, -12)$ into the original equations, to find and confirm the exact $x$ and $y$,
$$\begin{align*}
-12 x -3y &= 66 &\implies && 4x+y &= -22\\
-3x -12y &= 144 &\implies && x+4y &= -48
\end{align*}\\
\boxed{x = -\frac83} (\approx -2.67),\quad
\boxed{y = -\frac{34}3} (\approx -11.33)$$
Both $x$ and $y$ satisfy the original equations. Or similarly the other case by swapping $x$ and $y$.
A: 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.
