Question

Materi : Menyelesaikan sistem nonlinear dua variabel dengan mengubah ke bentuk sistem persamaan dua Variabel

Soal :
Tentukan penyelesaian dari sistem persamaan
berikut.
No. 4 dan 5

Answer 1

Verified answer

4.]
Misal √x = a , √(y + 5) = b

a - b = 3 |x 3|

3a - 3b = 9
4a + 3b = 17
__________ (+)
7a = 26
a = 26/7
√x = 26/7
x = 676/49

a - b = 3
b = a - 3
b = 26/7 - 3
b = 5/7
√(y + 5) = 5/7
y + 5 = 25/49
y = -220/49

HP : {(676/49 , -220/49)}

5.]
Misal 1/(x + 3) = a , 1/(y + 3) = b

a + 3b = 1
3a + b =1
_______ (–)
-2a + 2b = 0
2a = 2b
a = b

Maka
a + 3b = 1
b + 3b = 1
4b = 1
b = ¼
1/(y + 3) = ¼
y + 3 = 4
y = 1

a = b
a = ¼
1/(x + 3) = ¼
x + 3 = 4
x = 1

HP : { 1 , 1 }

Answer 2

4.]
Misal : √x = a , √(y + 5) = b

a - b = 3 |x 3|

3a - 3b = 9
4a + 3b = 17
__________ (+)
7a = 26
a = 26/7
√x = 26/7
x = 676/49

a - b = 3
b = a - 3
b = 26/7 - 3
b = 5/7
√(y + 5) = 5/7
y + 5 = 25/49
y = -220/49

HP : {676/49 , -220/49}

5.]
Misal : 1/(x + 3) = a , 1/(y + 3) = b

a + 3b = 1
3a + b =1
_______ (–)
-2a + 2b = 0
2a = 2b
a = b

Maka
a + 3b = 1
b + 3b = 1
4b = 1
b = ¼
1/(y + 3) = ¼
y + 3 = 4
y = 1

a = b
a = ¼
1/(x + 3) = ¼
x + 3 = 4
x = 1

HP : { 1 , 1 }


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