Question

Persamaan linear tiga variabel
jika
6/x+2 + 15/y+3 + 2/z+1= 8
4/x+2 + 5/y+3 + 3/z+1=6
8/x+2 - 10/y+3 + 5/z+1=5
tentukan nilai x+y+z

Answer 1

Misal: 1/(x+2) = a
          1/(y+3) = b
         1/(z+1) = c
sist persamaannya menjadi:
6a + 15b+ 2c= 8  ... pers 1
4a+ 5b + 3c=6  ... pers 2
8a - 10b + 5c=5 ... pers 3
eliminasi pers 1 dan 2
6a + 15b+ 2c= 8 (x1) --> 6a + 15b+ 2c= 8
4a+ 5b + 3c=6  (x3) --> 12a + 15b + 9c =18  _
                                       6a + 7c = 10 ... pers 4
eliminsai pers 2 dan 3
4a+ 5b + 3c=6 (x2) --> 8a + 10b + 6c = 12
8a - 10b + 5c=5 (x1) -->8a - 10b + 5c = 5  +
                                       16a + 11c = 17 ... pers 5
eliminasi pers 4 dan 5
6a + 7c = 10 (x8) --> 48a + 56c = 80
16a + 11c = 17 (x3) --> 48a + 33c = 51  _
                                           23c = 29
                                             c = 29/23
substitusi
6a + 7c = 10
6a + 7(29/23) = 10
6a +203/23 = 10
6a = 230/23 - 203/23
6a = 27/23
a = 9/46

8a - 10b + 5c=5
8(9/46) - 10b + 5(29/23) = 5
36/23 + 145/23 - 115/23 = 10b
66/23 = 10b
b = 33/115

maka
1/(x+2) = 9/46
9x+18 = 46
9x = 28
x = 28/9

1/(y+3) = 33/115
33y+99 = 115
33y = 16
y = 16/33

1/(z+1) = 29/23
29z+29 = 23
29z = -6
z = -6/29

x+y+z = 28/9 + 16/33 + -6/29
disamakan penyebut, dilanjutkan sendiri ya dek.. diteliti lagi angkanya.. agak rumit angkanya.. semoga membantu dan selamat belajar.. :)