Question

Soal persamaan 4 variabel

Level: Hard

ab + c + d = 3
bc + a + d = 5
cd + a + b = 2
da + b + c = 6

a + b + c + d = ...

Dilarang jawab secara asal :)

Solved it :)

Answer 1

Solusi :

ab + c + d = 3 

bc + a + d = 5 

cd + a + b = 2 

da + b + c = 6 

(1) + (2) = (3) + (4) à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) àbc + a + b = 5

Persamaan (3) àbc + a + b = 2

Kontradiksi maka tidak ada nilai a, b, c dan d yang memenuhi.

• Jika a + c = 2

(1) + (2) àab + bc + a + c + 2d = 8

b(a + c) + a + c + 2d = 8

b + d = 3

(2) + (3) àbc + cd + 2a + b + d = 7

c(b + d) + 2a + b + d = 7

3c + 2a = 4

3c + 2(2 − c) = 4 àc = 0 dana = 2

Persamaan (2) àb(0) + (2) + d = 5 àd = 3                                       

Persamaan (3) à0.3 + 2 + b = 2     àb = 0

(a, b, c, d) yang memenuhi adalah (2, 0, 0, 3).

Jadi a+b+c+d = 5.

Answer 2

Misal a+b+c+d = x
maka, ke-empat pers di atas dpt disajikan sbgai :
ab - (a+b) = 4 - x atau (a-1)(b-1) = 4 - x  ... (1)
bc - (b+c) = 6 - x atau (b-1)(c-1) = 6 - x  ... (2)
cd - (c+d) = 3 - x atau (c-1)(d-1) = 3 - x  ... (3)
da - (a+d) = 7- x atau (d-1)(a-1) = 7  - x ... (4)

pers (1) dikali dg pers (3), sdgkan Pers (2) dikali pers (4), get :
(a - 1)(b - 1)(c - 1)(d - 1) = (4 - x)(3 - x) dan
(a - 1)(b -1)(c - 1)(d - 1) = (6 - x)(7 - x)
akibatnya,
(4 - x)(3 - x) = (6 - x)(7 - x)
x² - 7x + 12 = x² - 13 x + 42
6x = 30
x = 5

berarti, nilai a + b + c + d = 5