Jawaban:

3. 2x + 3y = 12 ... (1)

3x + 2y = 8 ... (2)

Eliminasi y pers 1 dan 2

2x + 3y = 12 |x2| 4x + 6y = 24

3x + 2y = 8 |x3| 9x + 6y = 24 -

5x = 0

X = 0

Substitusi x=0 ke pers 1

2x +3y = 12

2(0) + 3y = 12

0 + 3y = 12

3y = 12

y = 4

Nilai a + b = 0 + 4 = 4

Jawabannya D. 4

4. y = 4x - 11

4x - y = 11 ... (1)

3y = -2x - 5

-2x -3y = 5 ... (2)

Eliminasi y pers 1 dan 2

4x - y = 11 |x3| 12x - 3y = 33

-2x - 3y = 5 |x1| -2x - 3y = 5 -

14x = 28

x = 28/14

X = 2

Substitusi x=2 ke pers 1

4x - y = 11

4(2) - y = 11

8 - y = 11

-y = 11 - 8

-y = 3

y = -3

Hp{2, - 3}

Jadi, titik potongnya adalah c. (2,-3)