Question

BIG POINT (50 POINT)

x + 2y = 12 ... (1)
3y + 3z = 9 ... (2)
2x + y + 2z = 16 ... (3)
Tentukan HP!!​

Answer 1

Verified answer

Jawaban:

Hp = {(6,4), (2,8), (0,2)}

Penjelasan dengan langkah-langkah:

Eliminasi persamaan 3 dan 1

2x + y + 2z = 16

x + 2y = 12

4x + 2y + 4z = 32

x + 2y = 12

dikurangi

3x + 4z = 20 ... (4)

Eliminasi persamaan 3 dan 2

2x + y + 2z = 16

3y + 3z = 9

6x + 3y + 6z = 48

3y + 3z = 9

dikurangi

6x + 3z = 39 ... (5)

Eliminasi persamaan 5 dan 4

3x + 4z = 20

6x + 3z = 39

6x + 8z = 40

6x + 3z = 39

dikurangi

5z = 1

z = 1/5 (0.2)

substitusi z ke persamaan 4

3x + 4/5 = 20

3x = 20-0,8

3x = 19,2

x = 64/10 (6,4)

substitusi x dan z ke persamaan 3

12,8 + y + 0,4 = 16

y = -12,8 -0,4 + 16

y = 28/10 (2,8)

x=6,4

y=2,8

z=0,2

Answer 2

Jawaban:

HP {x, y z} = {32/5, 14/5, 1/5}

Penjelasan dengan langkah-langkah:

x + 2y = 12 ... (1)

3y + 3z = 9 ... (2)

2x + y + 2z = 16 ... (3)

..

x + 2y = 12

x = 12 - 2y ... (4)

..

3y + 3z = 9

3y + z = 3

z = 3 - y ... (5)

..

Sub (4) dan (5) → (3)

2x + y + 2z = 16

2(12 - 2y) + y + 2(3 - y) = 16

24 - 4y + y + 6 - y = 16

5y = 14

y = 2,8 ... (Y)

..

Sub (Y) → (1)

x + 2y = 12

x + 2(2,8) = 12

x + 5,6 = 12

x = 6,4 ... (X)

..

Sub (Y) → (2)

3y + 3z = 9

3(2,8) + 3z = 9

8,4 + 3z = 9

3z = 0,6

z = 0,2

..

x = 6,4 = 6 + 2/5 = 32/5

y = 2,8 = 2 + 4/5 = 14/5

z = 0,2 = 1/5

[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at}}[/tex]