Tentukan nilai k untuk setiap persamaan garis berikut untuk: a. g1 : (3k)x-6y=20 dengan gradien sama dengan gradien g2 : x-2y=16. b. l1 : (3k+2)x+ky= 12 dengan

a) 6y = 3kx - 20
y = (k/2)x - 10/3
2y = x - 16
y = x/2 - 8
gradien sama maka:
k/2 = 1/2
k = 1

b) ky = - (3k + 2)x + 12
y = - (3k + 2)x/k + 12/k
(k - 6)y = 7x + 16
y = 7x/(k - 6) + 16/(k - 6)
gadien sama maka:
- (3k + 2)/k = 7/(k - 6)
- (3k² - 18k + 2k - 12) = 7k
3k² - 16k - 12 = - 7k
3k² - 9k - 12 = 0
k² - 3k - 4 = 0
(k + 1)(k - 4) = 0
k = - 1 atau k = 4

a. gradien g2
 2y = x -16
 y = 1/2 x -8
gradiennya = 1/2
gradien g1
(3k)x - 6y = 20
6y = 3kx -20
y = 3k/6 x- 20/6
gradien g1 3k/6
3k/6 = 1/2
6k = 6
 k = 1
b. gradien l2
(k-6)y = 7x +16
y = 7/(k-6) + 16/ (k-6)
gradiennya = 7/(k-6)
gradien l1
ky = - (3k+2) x +12
y = -(3k+2)/k + 12/k
gradiennya = -(3k+2)/k
-(3k+2)/k = 7/(k-6)
(3k+2)(k-6) = -7k
3k^2 - 16k -12 = -7k
3k^2 - 9k - 12 = 0
k^2 - 3k - 4 =0
(k-4)(k+1)
k = 4 atau k = -1