Solve for $x$ in the equation containing ${\lfloor{x}\rfloor}$ and $\{x\}$
hint
Observe that
$$\lfloor x-2 \rfloor =\lfloor x\rfloor -2$$
the equation becomes
$$\lfloor x \rfloor=2\{x\}+4$$
Decompose $x=n+f$. Then
$$\frac n{n-2}-\frac{n-2}n=\frac{8f+12}{n(n-2)}$$ or
$$n^2-(n-2)^2=8f+12$$
or
$$f=\frac{n-4}2.$$
The possible values for $n$ are $4$ and $5$ (so that $0\le f<1$).