Find all common solutions (mod 12) (or show that there are none to 4x+y≡6(mod 12), and x+4y≡2(mod 12)?
i know the answer is none but how to do i show the proof behind it?
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i know the answer is none but how to do i show the proof behind it?
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4x+y≡6(mod 12) (1)
x+4y≡2(mod 12) (2)
multiply both sides of (2) by 4, counting 4×4=16≡4 (mod 12)
(x+4y)×4≡2×4
4x+4y≡8 (3)
(4x+4y)-(4x+y)≡8-6 ....(3)-(1)
3y ≡ 2
but 3y∈ { 0,3,6,9 }
There is no answer to the problem.