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Misal :
1/x = a
1/y = b
1/z = c

a + b + c = 6 … P1
2a + 2b – c = 3 … P2
3a – b + 2c = 7 … P3

Eliminasi P1 dan P2
a + b + c = 6 |x 2|
2a + 2b – c = 3 |x 1|
—————————
2a + 2b + 2c = 12
2a + 2b – c = 3
————————- –
3c = 9
c = 3

Eliminasi P1 dan P3
a + b + c = 6
3a – b + 2c = 7
——————— +
4a + 3c = 13
4a + 3(3) = 13
4a + 9 = 13
4a = 13 – 9
4a = 4
a = 1

Substitusi a dan c ke P1
a + b + c = 6
1 + b + 3 = 6
b + 4 = 6
b = 6 – 4
b = 2

1/x = a
1/x = 1
x = 1

1/y = b
1/y = 2
y = 1/2

1/z = c
1/z = 3
z = 1/3

maka :
x + 2y + 3z
= 1 + 2(1/2) + 3(1/3)
= 1 + 1 + 1
= 3

Jawaban C