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: