Solving $7[x]+23\{x\}=191$ 
For every real number $x$, $[x]$ denotes the largest integer less than
  or equal to $x$ and $\{x\}=x-[x]$. 
The number of real solutions of 
$$7[x]+23\{x\}=191$$ is
(a) 0 $\quad$ (b) 1 $\quad$ (c) 2 $\quad$ (d) 3

I solved it like this:
$$7[x]+23\{x\}=191$$
$$7(x-\{x\})+23\{x\}=191$$
$$7x+16\{x\}=191$$
Now,
$$0 \leq \{x\} < 1$$
$$0 \leq 16\{x\} < 16$$
$$0 \leq 191-7x < 1$$
$$-16 < 7x-191 \leq 0$$
$$175 < 7x \leq 191$$
$$\frac{175}{7} < x \leq \frac{191}{7}$$
$$25 < x \leq 27.2857$$
But $x=26$ doesn't satisfy given equation. Answer given is $(d)$ $3$.
I don't understand where I'm wrong.  
A hint will be of great help.
Please don't provide complete solution.
Sorry for poor LaTeX.
 A: $$0\le\{x\}<1\implies0\le23\{x\}<23$$
$$\implies0\le191-7[x]<23\iff0\ge7[x]-191>-23$$
$$\iff28>191/7\ge[x]>(191-23)/7=24$$
So, $[x]$ can be $25,26,27$
A: Fractional part of x , $\{x\}$ must be of the form $\frac{y}{23}$. Thus, write $x = z + \frac{y}{23}$ where $z = [x]$ and $y<23$ and $y,z$ are integers. Thus, $7z + y = 191$. 
$$191 \equiv 2 \mod 7$$
Thus, $y \equiv 2 \mod 7$. Number of such values of $y$ that are less than 23 are just 3 (2,9,16).
A: You should work with $[x]$ and not $x$, try
$$0≤ \{x\} <1$$
$$0≤ 23\{x\} <23$$
$$0≤191−7[x]<23$$
$$−23<7[x]−191≤0$$
$$168<7[x]≤191$$
$$168/7<[x]≤191/7$$
$$24<[x]≤27.2857$$
So your answers are some $x$ with $[x]$ = $25$, $26$ or $27$
The $\{x\}$ part is just $\frac{(191 - 7[x])}{23}$ 
then :)
A: This is my simple solutions,
Hope it's help
$\mathrm{7}\left[{x}\right]+\mathrm{23}\left\{{x}\right\}=\mathrm{191} \\ $
$\left[{x}\right]=\mathrm{27}+\frac{\mathrm{2}−\mathrm{23}\left\{{x}\right\}}{\mathrm{7}}\:\:\:\:\:\:\:\:\:\: \\ $
$\frac{\mathrm{2}−\mathrm{23}\left\{{x}\right\}}{\mathrm{7}}={k}\in{Z} \\ $
$\frac{\mathrm{2}−\mathrm{7}{k}}{\mathrm{23}}=\left\{{x}\right\}\in\left[\mathrm{0},\mathrm{1}\right) \\ $
${k}\in\left\{\mathrm{0},-\mathrm{1},-\mathrm{2}\right\}\Rightarrow{n}\left({k}\right)=\mathrm{3} \\ $
