Soal Olimpiade Matematika 2

Selesaikan persamaan simultan : ab + c + d = 3, bc + a + d = 5, cd + a + b = 2, da + b + c = 6 dengan a, b , c dan d adalah bilangan real.
(Sumber : British Mathematical Olympiad 2003/2004 Round 1)

Solusi :
ab + c + d = 3 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ (1)
bc + a + d = 5 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ (2)
cd + a + b = 2 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ (3)
da + b + c = 6 ⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅ (4)
(1) + (2) = (3) + (4) -> ab + c + d + bc + a + d = cd + a + b + da + b + c b(a + c) + 2d = d(a + c) + 2b (b − d)(a + c) = 2(b − d) (b − d)(a + c − 2) = 0 b = d atau a + c = 2
• Jika b = d , Persamaan (2) -> bc + a + b = 5, Persamaan (3) -> bc + a + b = 2
Kontradiksi maka tidak ada nilai a, b, c dan d yang memenuhi.
• Jika a + c = 2, (1) + (2) -> ab + bc + a + c + 2d = 8 b(a + c) + a + c + 2d = 8 b + d = 3
(2) + (3) -> bc + cd + 2a + b + d = 7 c(b + d) + 2a + b + d = 7 3c + 2a = 4 3c + 2(2 − c) = 4
-> c = 0 -> a = 2
Persamaan (2) -> b(0) + (2) + d = 5 -> d = 3 -> b = 3 − (3) = 0
(a, b, c, d) yang memenuhi adalah (2, 0, 0, 3)

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