Multiplying a linear congruence by an integer that is not coprime with modulo $n$ can result in a different solution set. It seems like such an operation is called an noninvertible transformation.
However, in the following example, the solution set appears to be different from the original even though i did not multiply the equation by an integer.
Since $\gcd(7,39)=1$, the equation $$\tag{17}7x\equiv22\pmod{39}\implies x\equiv31\pmod{39}$$ has a unique solution $(\bmod{39})$. Subtracting equation (17) five times from $$39x\equiv0\pmod{39}$$ gives $$4x\equiv-110\equiv-110+3\cdot39\equiv7\pmod{39}$$ We subtract this from (17) and get $$\tag{18}3x\equiv15\pmod{39}$$ Here we have the opportunity to divide both sides by $3$. But since $\gcd(3,39)=3$, $$x\equiv5\pmod{13}$$ This is equivalent to (3 solutions) $$x\equiv5,18,31\pmod{39}$$
This example is taken from An Introduction to Number Theory - Harold M. Stark
My question is why the solution set changed even though I didn't do the multiplication. What types of operations on linear congruences yield extraneous solutions? As you can see from the examples above, it seems that multiplying by an integer that is not coprime to the modulo is not the only reason.
