soal madas um undip 2012, tolong bantuannya :)

3/x + 4/y + 5/z = 4
5/x - 6/y + 1/z = 17/30
6/x + 1/y - 3z = 21/20

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

3a + 4b + 5c = 4 (× 1)
5a - 6b + c = 17/30 (× 5)

...3a + 4b + 5c = 4
25a - 30b + 5c = 17/6
--------------------------------- -
-22a + 34b = (24-17)/6
-22a + 34b = 7/6 => pers.1

3a + 4b + 5c = 4 (× 3)
6a + b - 3c = 21/20 (× 5)

...9a + 12b + 15c = 12
30a + ..5b - 15c = 21/4
--------------------------------- +
39a + 17b = (48+21)/4
39a + 17b = 69/4 => pers.2

-22a + 34b = 7/6 (× 1)
39a + 17b = 69/4 (× 2)

-22a + 34b = 7/6
..78a + 34b = 69/2
----------------------------- -
-100a = (7-207)/6
-100a = -200/6
a = (-200/6)/-100
a = (-200/6)×(-1/100)
a = 200/600
a = 1/3

-22a + 34b = 7/6
-22(1/3) + 34b = 7/6
-22/3 + 34b = 7/6
34b = 7/6 + 22/3
34b = 7/6 + 44/6
34b = 51/6
b = (51/6)/34
b = (51/6)×(1/34)
b = 51/204
b = 1/4

3a + 4b + 5c = 4
3(1/3) + 4(1/4) + 5c = 4
1 + 1 + 5c = 4
2 + 5c = 4
5c = 4 - 2
5c = 2
c = 2/5

a = 1/x
1/3 = 1/x
x = 3

b = 1/y
1/4 = 1/y
y = 4

c = 1/z
2/5 = 1/z
z = 5/2

Hp = {3 , 4 , 5/2}
Jadi himpunan penyelesaian dari sistem persamaan diatas adalah {(3,4,5/2)} (D)