Chinese Remainder Thm/Simult. congruences: X ≡ 2 mod 3, X ≡ 2 mod 4, X ≡ 1 mod 5

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Chinese Remainder Thm/Simult. congruences: X ≡ 2 mod 3, X ≡ 2 mod 4, X ≡ 1 mod 5

"Find an integer X such that


X ≡ 2 mod 3,


X ≡ 2 mod 4, and

X ≡ 1 mod 5."


So this is what I did for the first two congruences:


x = 3a + 2


x = 4b +2


3a + 2 = 4b +2


Possible value for a is 1 and b is 3/4.


Therefore x = 3(1) + 2 = 4(3/4) + 2 = 5 (mod 12).


So x = 12c + 5


And the 4th congruence has equation:


x = 5d +1


12c + 5 = 5d + 1


5d - 12c = 4.


Possible value of d is 4 and c is 4/3.


12(4/3) + 5 = 5(4) + 1 = 21 (mod 60)


But 21 is clearly wrong.


It seems like I even got the calculation for the first two congruences wrong.


I have seen this method work before so I'm not sure where I've gone wrong here.


Any help?