Linear congruence fill in the missing step?

Currently working on this problem and I'm having trouble seeing how it goes from one line to the next.

$45x \equiv 63\mod 11$ goes to $x \equiv 8\mod 11$

Any help would be awesome thanks.

Please try to explain how it works to help me see thanks.

The intuition is that one can replace an argument of a sum or product by any congruent integer (not true for exponents!). This follows by the Polynomial Congruence Rule, i.e. by iterating the congruence Sum and Product rules. Let's do this very carefully for your example.

Note that $\ \ \color{#c00}{45\equiv 1}\pmod{11}\ $ by $\ 11\mid 45\!-\!1 = 44$

Therefore $\ \color{#c00}{45}x\equiv \color{#c00}1x\equiv x\,$ by the Congruence Product Rule.

Similarly $\ \ \color{#0a0}{63\equiv 8}\pmod{11}\ $ by $\ 11\mid 63\!-\!8 = 55.\ $ Chaining together these congruence equations

$\qquad x\equiv \color{#c00}1x\equiv\color{#c00}{45}x \equiv \color{#0a0}{63 \equiv 8}\,$ so $\,x\equiv 8\,$ since $\,\equiv\,$ is transitive, being an equivalence relation.