The congruence has a solution if and only if d=gcd(a,m) divides b. The solution is given by . That much I know. And I see you have put the system into upper triangular form, which is probably a good idea.

One idea is to break the modulus down to powers of primes (in your case 228=4*3*19) and solve for each modulus, then use the Chinese Remainder Theorem to get the final solution. I haven't worked out what happens for your system.

I would just work backward through the equations (as you did):

, since gcd(226,228)=2. Or,

where z=0 or 1. Now the next equation up:

(z dropped out of the equation, otherwise we would treat it like a constant)

where w=0,1, or 2. So that gives us six solutions. But there is another equation:

Since gcd(224,228)=4 and gcd(225,228)=1, substitute for y:

and since gcd(224,228)=4 must divide 142-114z, we must have z=1.

Now substitute for x:

and only w=1 works. So only the solution z=1, w=1 is valid, so the solution is:

- Hollywood