For example: Solve for $a$ given the following linear congruence.
$$ \begin{align}x &≡ a \mod 7 \\x &≡ 3 \mod 35\end{align}$$
So far I tried solving the question using the Chinese remainder theorem mainly by solving "backwards" but I just can't seem to make it work
$\ \qquad\qquad\qquad\quad\begin{align}\bmod 5(\color{#c00}7)\!:\ &3\equiv x\\ \color{#0a0}\Longrightarrow\ \bmod\ \ \ \color{#c00}7\!:\ \ \ &3\equiv x\equiv a\iff a = 3+7n\ \ \text{for some integer $n$}\end{align}$
by congruences $\,\rm\color{#0a0}{persist}\,$ mod $\rm\color{#c00}{factors}$ of the modulus, by here. This fundamental property of congruences should be known like the back of your hand to be proficient with congruences.
translate the modular statements to divisibility statements:
if $7\times5=35|x-3$ then certainly $7|x-3$, so $a=3$
Given $$x\equiv a \mod 7 \implies x = 7k +a \quad \text{ (1.)}$$
Also $$x \equiv 3 \mod 35 \implies x = 35l + 3 \quad \text { (2.)}$$
multiplying $(1.)$ by $5$ and subtracting $(2.)$ from it ,
$$5x -x = 35(k-l) + (5a-3) \\ \color{#0c9}{4x \equiv \,(5a-3) \mod 35}$$
But from $(2.)$ , we also get $\,\,\,4x \equiv 12 \mod 35 $.
By comparing the equations , we get
$$5a-3 = 12 \implies \boxed{\color{#2ca}{a = 3}}$$
