Question

Tentukan himpunan penyelesaian sistem persamaan berikut dengan metode substitusi dan eliminasi
1. a + 2b = -5 dan 3a - 2b = 1
2. 4x + 3y = 47 dan 3x + 2y = 34

Answer 1

1. a + 2b = -5 dan 3a - 2b = 1
a = -5 - 2b substitusi ke 3a - 2b = 1
3a - 2b = 1
3(-5 - 2b) - 2b = 1
-15 - 6b - 2b = 1
-15 - 8b = 1
-8b = 1 + 15
-8b = 16
b = 16/-8
b = -2

b = -2 substitusi ke a + 2b = -5
a + 2b = -5
a + 2(-2) = -5
a - 4 = -5
a = -5 + 4
a = -1

HP = {-1,-2}

2. 4x + 3y = 47 dan 3x + 2y = 34
y = (34 - 3x)/2
y = 17 - 3x/2 substitusi ke 4x + 3y = 47 
4x + 3y = 47 
4x + 3(17 - 3x/2) = 47 
4x + 51 - 9x/2 = 47
8x - 9x/2 = 47 - 51
-1/2x = -4
x = -4 : -1/2
x = -4 × 2/-1 = -4 × -2
x = 8

x = 8 substitusi ke 4x + 3y = 47
4x + 3y = 47
4(8) + 3y = 47
32 + 3y = 47
3y = 47 - 32
3y = 15
y = 15/3
y = 5

HP = {8,5}

Answer 2

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

2) 4x+3y=47 |×3| 12x+9y=141
3x+2y=34 |×4| 12x+8y=136
----------------------------------(-)
y = 5
4x + 3y = 47
4x + 3.5 = 47
4x + 15 = 47
4x = 47 - 15
4x = 32
x = 8