Easier way to compute the solution to a modular equation system

By the Easy CRT Formula, & by twiddling to an $\color{#c00}{\text{exact quotient}\equiv 5}$

$\ \begin{align}\rm x&\equiv\color{#0a0}{5\!\!\!\pmod{\!70}} \\ \rm x&\equiv \color{#90f}{4\!\!\!\pmod{\!39}}\end{align} \!\!\iff\! $ $ x \equiv\, \color{#0a0}{5\! +\! 70} \bigg(\!\underbrace{\dfrac{\color{#90f}4\!-\!\color{#0a0}5}{\color{#0a0}{70}}\!\color{#90f}{\bmod{39}}}_{\small{\dfrac{\color{#90f}{-}\color{#0a0}1}{\color{#0a0}{70}}\ \,\equiv\,\ \color{#c00}{\dfrac{-40}{-8}\ \equiv\ \large 5}}}\!\bigg)$ $\equiv 355\pmod{\!\color{#0a0}{70}\cdot \color{#90f}{39}}$

let $x=39y+4$
$39y+4\equiv 5\pmod{70}$
$39y\equiv 1\pmod{70}$
the inverse of 39 mod 70 is 9
$y\equiv 9\pmod{70}$
$y=70z+9$
$x=39(70z+9)+4=39*70z+355$
$x\equiv 355\pmod{39*70}$

This can be done entirely in your head.

From the second congruence, we know that $x=5+70n$ for some $n$. Thus we seek to solve $$5+70n\equiv 4\pmod {39}$$

But this is equivalent to $$8n\equiv 1 \pmod {39}\implies n\equiv 5\pmod {39}$$ and we are done.