Question

for which value k , 3x-y+8=0 and 6x-ky=-16 have infinitely many solution s?

Answer 1

HEY MATE HERE IS YOUR ANSWER

For the lines to be coincident

$\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2} - - - - - > (1)$
3x-y+8=0 6x-ky+16=0

a1x+b1+c1=0 a2x+b2y+c2=0

COMPARE THEM

a1=3 b1=-1 c1=8 a2=6 b2=-k c2=16

substitute them in eq 1

$\frac{3}{6} = \frac{ - 1}{ - k} = \frac{8}{16} \\ \frac{1}{2} = \frac{1}{k} = \frac{1}{2} \\ k = 2$
HOPE IT HELPS YOU

FOLLOW ME FOR ANY OTHER QUESTIONS

Answer 2

Answer:

k=3

Step-by-step explanation:

For infinite solution 39=−1−k=824

⇒13=1k⇒k=3