Diketahui bentuk al jabar berikut: (X-1) (X+1) (X+2)=ax3+bx2-cx-d jika ax+by=4 ex=6,e=c+d tentukan nilai y?

(x - 1) (x + 1) (x + 2)
= (x² - 1) (x + 2)
= x³ + 2x² - x - 2  =  ax³ + bx² - cx - d
berarti
a = 1
b = 2
c = 1
d = 2

diketahui 
ax + by = 4
1x + 2y = 4 ......................(1)

e = c + d
e = 1 + 2
e = 3

ex = 6
3x = 6
x = 6/3
x = 2

subtitusikan x = 2 ke pers. (1)
1x + 2y = 4
1(2) + 2y = 4
2y = 4 - 2
2y = 2
y = 2/2
y = 1

jadi nilai y = 1