Jawab:

a. x = 12 dan y = 1

b. x = -3 dan y = -1

c. x = 4 dan y = -4

d. x = 7 dan y = 14

Penjelasan dengan langkah-langkah:

a. 6x - 5y = 67 dan x + 3y = 15

-> metode eliminasi

6x - 5y = 67

x + 3y = 15

-------------------- (di setarakan pada bagian x)

6x - 5y = 67

6x + 18y = 90

----------------------- -

-23y = -23

y = 1

masukkan (subtitusi) nilai y = 1 ke persamaan (2)

x + 3y = 15

x + 3 (1) = 15

x = 12

b. 2 (x-2y) = -2 dan 4x = 6x + 6

-> metode subtitusi

4x = 6x + 6

4x - 6x = 6

-2x = 6

x = -3

masukkan (subtitusi) nilai x = -3 ke persamaan (1)

2 (x-2y) = -2

2x - 4y = -2

2(-3) - 4y = -2

-4y = 4

y = -1

c. 2x + 3y = -4 dan x - 5y = 24

-> metode eliminasi

2x + 3y = -4

x - 5y = 24

-------------------- (di setarakan pada bagian x)

2x + 3y = -4

2x - 10y = 48

-------------------- -

13y = -52

y = -4

masukkan (subtitusi) nilai y = -4 ke persamaan (1)

2x + 3y = -4

2x + 3(-4) = -4

2x = -4 + 12

2x = 8

x = 4

d. 1/2y = 2x -7 dan 5x - 3y + 7 = 0

1/2y = 2x -7

y = 4x - 14 .... (3)

masukkan (subtitusi) y = 4x - 14 ke persamaan (2)

5x - 3y + 7 = 0

5x - 3(4x - 14) + 7 = 0

5x - 12x + 42 + 7 = 0

-7x = -49

x = 7

masukkan (subtitusi) x = 7 ke persamaan (2)

y = 4x - 14

y = 4(7) -14

y = 28 -14

y = 14