What are all these brackets?

$7 \lfloor x \rfloor + 23 \{ x \} = 191 \quad \Rightarrow \{ x \} = \dfrac {191- 7 \lfloor x \rfloor} {23} \quad \Rightarrow 0 < \dfrac {191- 7 \lfloor x \rfloor} {23} < 1 \\ \Rightarrow 0 < 191- 7 \lfloor x \rfloor < 23 \quad \Rightarrow 168 <7 \lfloor x \rfloor < 191 \quad \Rightarrow \dfrac{168}{7} < \lfloor x \rfloor < \dfrac {191}{7} \\ \Rightarrow 24 < \lfloor x \rfloor < 27.285 \quad \Rightarrow 25 \le \lfloor x \rfloor \le 27$

Therefore, the number of solutions is $\boxed{3}$ .

$23\{ x\} =191-7\left\lfloor x \right\rfloor \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (1)\\ Since\quad RHS\quad is\quad an\quad integer,\quad LHS\quad is\quad also\quad an\quad integer.\\ So\quad 23\{ x\} \quad is\quad an\quad integer.\\ We\quad know\quad that\quad 0\le \{ x\} <1\\ So\quad \{ x\} \quad is\quad in\quad the\quad form\quad \frac { a }{ 23 } \quad where\quad a\quad is\quad a\quad whole\quad number\quad from\quad 0\quad to\quad 22.\\ So\quad 23\{ x\} \quad is\quad a\quad whole\quad number\quad from\quad 0\quad to\quad 22.\\ Rearranging\quad (1),\quad we\quad get\\ 7\left\lfloor x \right\rfloor =191-23\{ x\} \\ LHS\quad is\quad always\quad a\quad multiple\quad of\quad seven,\quad so\quad we\quad just\quad need\quad to\quad check\quad \\ how\quad many\quad multiples\quad of\quad 7\quad lie\quad from\quad 191-22\quad (=169)\quad to\quad 191.\\ \\ This\quad comes\quad out\quad to\quad be\quad \boxed { 3 }$

Did the same way :-)