Computing linear congruence

Question

I have the following linear congruence:

$$40x\equiv 28\pmod{73}$$

I have attempted to simplify the congruence by dividing both terms by 4,

$$10x \equiv 7\pmod{73}$$

but I can't figure out what to do next.

What would be the best method around solving this?

Answer 1

Note that $10$ and $73$ are relatively primes thus by Bezout Identity we can find by Euclidean algorithm $a$ and $b$ such that

$$10a+73b=1\implies 10 a =1-73b$$

thus $a$ is the inverse of $10 \mod{73}$ and then

$$10x \equiv 7 \mod {73}\implies 10ax \equiv 7a \mod {73}\implies x \equiv 7a \mod {73}$$

Answer 2

We get $$x\equiv \frac{7}{10}\equiv \frac{80}{10}\equiv 8 \mod 73$$