Question

Find the value of k for which the system of equations have infinitely many solutions x-ky=2 & 3x+6y=-5

Answer 1

Hi mate

Here is ur answer

X - ky = 2

x - ky - 2 = 0.............(1)

3x + 6y = -5

3x + 6y + 5=0............(2)

Infinitely many solution

so,

a1/a2 = b1/b2 = c1/c2

1/3 = -k/6 = -2/5

1/3 = -k/6

2 = -k

k=(-2)

-k/6 = -2/5

-k = -12/5

k=12/5

i hope this will help u........

Mark as brainliest

Answer 2

Answer:

k = 5

Step-by-step explanation:

x – ky = 2

3x + 6y = –5

For infinitely many solutions:-

a1/a2 = b1/b2 = c1/c2

ie, a1 = 1

b1 = –k

c1 = 2

and

a2 = 3

b2 = 6

c2 = –5

Therefore,

a1/a2 = 1/3

b1/b2 = –k/6

c1/c2 = 2/–5

ie, 1/3 = k/6 = 2/–5

1/3 = –k/6

ie,

6 = –3k = 2/ –5

k = 6/–3, = 2/ –5

k = –2, = 2/–5

k = 10/2

ie,

k = 5

THANK YOU