Dividing the linear congruence equations

Solution 1:

Consider the congruence $$ax\equiv ay \pmod{n}.\tag{1}$$

If $a$ and $n$ are relatively prime, then the congruence (1) is equivalent to the congruence $x\equiv y\pmod{n}$.

More generally, if $\gcd(a,n)=d$, then the congruence (1) is equivalent to the congruence $x\equiv y \pmod{\frac{n}{d}}$. If we put $d=1$, we arrive at the special case dealt with in the preceding paragraph.

For your particular numbers, the original congruence is equivalent to $7x\equiv 2\pmod{15}$. This has the solution $x\equiv 11\pmod{15}$. This solution is unique modulo $15$.

If we want to express the answers modulo the original $90$, there are several solutions. For $x\equiv 11\pmod{15}$ if and only if $x$ is congruent to one of $11,26,41,56,71,86$ modulo $90$. (We keep adding $15$'s.)