A=5(mod 12) and b=11(mod 8); what is the remainder of (7a+b)/4

I understand how to add this with the same modulo but how does this work?

12 | (a-5) implies 4|(a-5) ==> 4|(a-1)==> a = 1mod4
8|(b-11)==>4|(b-11)==>4|(b-3)==> b = 3mod4
So, 7a + b = 7(1) + (3) = 10 = 2 mod4.

Sorry about the symbols, a==>b means a implies b
a|b means a divides b. My idea was to get the mod4 information
out of the mod12 and mod8 information because without that
the problem wouldn't solve. Basically if 12 divides something
it brings the 4 with it since 4 divides 12 etc. same with 8.

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a = 5x12 = 60, b = 88

7x60+88 = 508

508/4 = 127.

See BODMOS rule