How to solve $(x \mod 7) - (x \mod 8) = 5$? I'm trying to solve $(x \mod 7) - (x \mod 8) = 5$ but no idea where to start. Help appreciated!

If CRT = Chinese Remainder Theorem is known then we can reformulate it as 
$$\begin{align}
x&\equiv a\!+\!5\!\!\pmod{7},\ \ \ \overbrace{ 0\le a\!+\!5 \le 6}^{\Large \iff a\  =\ 0,1}\\
x&\equiv a\quad\pmod{8},\ \ \  0\le a\le 7\end{align}\qquad \qquad\qquad $$
Solve the $\,a=0\,$ case $\,x\equiv (5,0)\bmod (7,8),\,$ then $\,x\!+\!1\equiv (6,1)\,$ yields the $\,a=1\,$ case.

W/o $ $  CRT: $\,\ \overbrace{ x-7j}^{\large x\bmod 7}-(\overbrace{x-8k)}^{\large x\bmod 8} = 5\iff 8k - 7j = 5\,\Rightarrow\, j \equiv \color{#c00}{5}\pmod{\!8},\ k\equiv \color{#c00}{5}\pmod{\!7}$
Working in the intial range $\, 0\le x\le 55\ $ and enforcing the remainder bounds
$\qquad\quad  \begin{align} 0\le \overbrace{x-7\cdot \color{#c00}5}^{\large x\bmod 7} \le 6\iff 35\le x \le \color{#0a0}{41}\\[.5em] 
0\le \underbrace{x-8\cdot \color{#c00}5}_{\large x\bmod 8} \le 7\iff \color{#0a0}{40}\le x \le 47\end{align}$
Therefore $\, x\equiv \color{#0a0}{40},\color{#0a0}{41}\pmod{56}$
