a. Dengan metode substitusi x + y = 2 y = 2 – x 2x + 3y = 6 2x + 3 (2 – x) = 6 2x + 6 – 3x = 6 -x = 0 x = 0 y = 2 – x y = 2 – 0 y = 2 hp = (x,y) (0,2); x,y Є Real
b. Dengan metode eliminasi 2x + 3y = 6 x1 2x + 3y = 6 x + y = 2 x(-2) -2x – 2y = -4 + y = 2 x + y = 2 x + 2 = 2 x = 0 hp = (x,y) (0,2); x,y Є Real
2x + 3y = 6 a1b2 – b1a2 = (2)(1) – (3)(1) = -1 ≠ 0, x + y = 2 maka memiliki penyelesaian. hp = (x,y) (0,2); x,y Є Real
a. Dengan metode substitusi
x + y = 2 y = 2 – x
2x + 3y = 6
2x + 3 (2 – x) = 6
2x + 6 – 3x = 6
-x = 0
x = 0
y = 2 – x
y = 2 – 0
y = 2
hp = (x,y) (0,2); x,y Є Real
b. Dengan metode eliminasi
2x + 3y = 6 x1 2x + 3y = 6
x + y = 2 x(-2) -2x – 2y = -4 +
y = 2
x + y = 2 x + 2 = 2 x = 0
hp = (x,y) (0,2); x,y Є Real
2x + 3y = 6 a1b2 – b1a2 = (2)(1) – (3)(1) = -1 ≠ 0,
x + y = 2 maka memiliki penyelesaian.
hp = (x,y) (0,2); x,y Є Real