Find all solutions of the equation $13[x]+25\{x\}=271$ 
For a real number $x$, let $[x]$ denote the largest integer $\le x$, and let $\{x \}$ denote $x-[x]$. Find all solutions of the equation:
$$13[x]+25\{x\}=271.$$

I tried to simplify the equation by:
$$13[x]+25(x-[x])=271$$
$$\implies 13[x]+25x-25[x]=271$$
$$\implies 25x-12[x]=271$$
$$\implies x= \dfrac{271+12[x]}{25}.$$
Now I saw that for $[x]=17$, we get $x=19$, but this is not a solution.
Do I have to go on manually checking for such solution, if it exists? Is there any generic/empirical way to solve this?
Some more facts I have observed (I don't know whether I am hitting at the right point or not).
If $x=[x]$,
$$13x=271$$
$$\implies x=\dfrac{271}{13}\approx 2.84.$$
Thus $[x]=2.$
But if $x>[x]$, then $$13[x]+25\{x\}=271,$$
where $\{x\}$ is the fractional part of $x$.
Thus $25\{x\}=271-[x]$.
But as $[x]$ is always an integer, $271-[x] \ne 25\{x\}.$
So we have only solution for $x=[x]$.
Possibly I am wrong at this. Please show me the correct way to solve it.

$\displaystyle \frac{25x-271}{12}=\lfloor x \rfloor$.
\begin{align*}
x-1<\frac{25x-271}{12}&\le x\\
19+\frac{12}{13}< x&\le20+\frac{11}{13}\\
\lfloor x \rfloor&=19 \textrm{ or }20
\end{align*}
If $\lfloor x \rfloor=19$, $\displaystyle x=\frac{12(19)+271}{25}=19.96$.
If $\lfloor x \rfloor=20$, $\displaystyle x=\frac{12(20)+271}{25}=20.44$
The only answers are $19.96$ and $20.44$.

The other way to solve this is to write $\displaystyle x=\frac{12\lfloor x\rfloor+271}{25}$. Then
\begin{align*}
\lfloor x\rfloor\le \frac{12\lfloor x\rfloor+271}{25}&<\lfloor x\rfloor+1\\
18+\frac{12}{13}<\lfloor x\rfloor&\le 20+\frac{11}{13}
\end{align*}
As $\lfloor x\rfloor$ is an integer, $\lfloor x\rfloor=19$ or $20$.
When $\lfloor x\rfloor=19$, $\displaystyle  x=\frac{12(19)+271}{25}=19.96$.
When $\lfloor x\rfloor=20$, $\displaystyle  x=\frac{12(20)+271}{25}=20.44$.
