Finding solutions of the system $27x + 90 \equiv 18 \pmod{99}$ I have to find solutions for the expression $$27x + 90 \equiv 18 \pmod{99}$$ 
My only problem is that I can only solve expressions like $ax \equiv  b \pmod{n}$.
How can I get rid of the $90$? Subtracting $90$ from both sides won't work I think because I will have a negative number...
Thanks a lot.
 A: You can always add or subtract multiples of 99 from either side.
$$27x +90 \equiv 18 \text{ mod 99}$$
$$27x +90-99 \equiv 18 \text{ mod 99}$$
$$27x -9\equiv 18 \text{ mod 99}$$
$$\text{ Now adding 9 to both sides gives,}$$
$$27x \equiv 27 \text{ mod 99}$$
$$\text{And dividing both sides by 9}$$
$$3x \equiv 3 \text{ mod 11}$$
$$\text{ And since 3 is coprime to 11}$$
$$x\equiv 1 \text{ mod 11}$$
$$\text{ From which one can see, $x=1$ is clearly a solution}$$
$$\text{ And for additional solutions just add multiples of 11}$$
A: Hint:
$$\quad 90\equiv -9\bmod 99\quad $$
A: Modulo $99$, there's no such thing as a negative number. $-100$, $-1$, $98$, and $197$ are all the same thing. Your idea of subtracting 90 from both sides is the thing to do.
A: Add 9 to both sides to get
$$27x + 99 \equiv 27{ \rm mod} 99,$$
so now you  aare down to 
$$27x\equiv 27{ \rm mod} 99,$$
One quick solution is $x = 1$.
A: $\rm mod\ 99\!:\ 27x\equiv 18\!-\!90\equiv 18\!+\!9\equiv 27\!\iff\! 99\mid 27(x\!-\!1)\!\iff\! 11\mid 3(x\!-\!1)\!\iff\! 11\mid x\!-\!1$
