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3x + y + 2z = 9 ...pers(1)
2x + 3y + z = 10 ...pers(2)
2x + 3z + 3z = 14 ...pers(3)
jawab : metode eliminasi dan substitusi
* eliminasi pers (1) & (2)
3x + y + 2z = 9 |×3|
2x + 3y + z = 10 |×1|
9x + 3y + 6z = 27
2x + 3y + z = 10
________________ -
7x + 5z = 17 ...pers(4)
* eliminasi pers (2) & (3)
2x + 3y + z = 10
2x + 3y + 3z = 14
________________ -
-2z = -4
z = -4/-2
z = 2
* substitusikan nilai z ke pers (4)
7x + 5z = 17
7x + 5(2) = 17
7x + 10 = 17
7x = 17 - 10
7x = 7
x = 1
* substitusikan nilai x dan z ke pers (1)
3x + y + 2z = 9
3(1) + y + 2(2) = 9
3 + y + 4 = 9
y + 7 = 9
y = 9 - 7
y = 2
jadi nilai :
x = 1
y = 2
z = 2
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mohon maaf jika jawabannya salah