Question

For what value of k the pair of linear equations kx-4y=3, 6x-12y=9 has infinite number of solutions?

Answer 1

Solution

Equation = kx-4y=3,

= 6x-12y=9

in infinite number of solution

this condition is apply

$\frac{a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}$

a 1 = k

b 1 = -4

c1 = -3,

a 2 = 6

b2 = -12

c2 = -9

put all these values we get :

$\frac{k}{6} = \frac{4}{12} = \frac{ - 3}{ - 9} \\ \\ \frac{k}{6} = \frac{4}{12} \\ \\ k \: = 2$

Answer 2

Given,

Two equations are:

kx - 4y = 3 -----(1)

6x - 12y = 9 ------(2)

let the value of x be 1

Substitute x in eq - 2

6(1) - 12y = 9

6 - 12y = 9

6 - 9 = 12y

-3 = 12y

y = -3/12 = -1/4

substitute y in eq - 1

k(1) - 4(-1/4) = 3

k + 1 = 3

k = 3 - 1

k = 2

Therefore, the value of k is 2.