Pembahasan Kuis
1) 2x + 3y = 1 … pers.(1)
3x + y = 5 … pers.(2)
Jawab :
Dari pers.(2) didapat : y = 5 – 3x … pers.(3).
lalu substitusikan pers.(3) ke pers.(1) :
2x + 3y = 1
2x + 3(5 – 3x) = 1
2x + 15 – 9x = 1
2x – 9x = 1 – 15
-7x = - 14
x = 2
Nilai x = 2 lalu disubstitusikan ke pers.(3) :
y = 5 – 3x
y = 5 – 3(2)
y = 5 – 6
y = - 1
Jadi, Himpunan Penyelesaian = { (2, - 1)}
2) 2x + 3y = 1 … pers.(1)
4x – 3y = 11 … pers.(2)
Jawab :
2x + 3y = 1
4x – 3y = 11 +
⇔ 6x = 12
⇔ x = 2
Harga x = 2 lalu substitusikan ke pers (1) :
2x + 3y = 1
⇔ 2(2) + 3y = 1
⇔ 4 + 3y = 1
⇔ 3y = 1 – 4
⇔ 3y = - 3
⇔ y = - 1
Jadi, HP = {( 2,-1)}
3) 4x + 2y = 2 … pers.(1)
x – 2y = 3 … pers.(2)
Jawab :
4x + 2y = 2
x – 2y = 3 +
⇔ 5x = 5
⇔ x = 1
Harga x = 1 lalu substitusikan ke pers (1) :
4x + 2y = 2
⇔ 4(2) + 2y = 2
⇔ 8 + 2y = 2
⇔ 2y = 2 – 4
⇔ 2y = -2
⇔ y = -1
Jadi, HP = {( 1,-1)}
4) x + 3y = 8 … pers.(1)
2x + y = 6 … pers.(2)
Jawab :
x + 3y = 8
x = 8 – 3y
Substitusikan nilai x ke persamaan (2)
⇔ 2(8 – 3y) + y = 6
⇔ 16 – 6y + y = 6
– 5y = –10
y = 2
Nilai y = 2 substitusikan ke pers (1) :
x + 3y = 8
⇔ x + 3(2) = 8
⇔ x + 6 = 8
⇔ x = 2
Jadi, HP = {( 2,2)}
5) x + 4y = 10 … pers.(1)
3x + 2y = 10 … pers.(2)
Jawab :
x + 4y = 10 |x3| 3x + 12y = 30
3x + 2y = 10 |x1| 3x + 2y = 10 –
10y = 20
y = 2
Harga y = 2 lalu substitusikan ke pers (1) :
x + 4y = 10
⇔ x + 4(2) = 10
⇔ x + 8 = 10
⇔ x = 10 – 8
⇔ x = 2
Jadi, HP = {( 2,2)}
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