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xy+y² = 0…..(1)
x-2y = 3…….(2)
dari (2)→ x = 2y+3 (substitusi pd (1))
(2y+3)y + y² = 0
2y² + 3y + y² = 0
3y² + 3y = 0
3y(y + 1) = 0
y = 0 atau y = -1
substitusi pd salahsatu
y = -1 → x - 2(-1) = 3
jadi yang memenuhi persamaan x-2y = 3
x = 1 (salahsatunya)
xy+y² = 0…..(1)
x-2y = 3…….(2)
dari (2)→ x = 2y+3 (substitusi pd (1))
(2y+3)y + y² = 0
2y² + 3y + y² = 0
3y² + 3y = 0
3y(y + 1) = 0
y = 0 atau y = -1
substitusi pd salahsatu
y = -1 → x - 2(-1) = 3
jadi yang memenuhi persamaan x-2y = 3
x = 1 (salahsatunya)