Question

For what value of k the pair of linear equation kx - 4y = 3
and 6x - 12y = 9 are consistent. ​

Answer 1

Let, x = 1

then,

6x - 12y = 9

6(1) - 12y = 9

6 - 12y = 9

-12y = 9 -6

-12y = 3

y = 3/-12

y = 1/-4

So, if x = 1 then y = 1/-4

kx -4y =3

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

k + 1 = 3

k = 3 - 1

k = 2

Therefore k = 2 to get infinite number of solutions

hope it helps you....✌️

Answer 2

Given:

kx - 4y = 3 ... Eq.1

6x - 12y = 9 ... Eq.2

Use Eq.2 to equate x and substitute to Eq.1

6x - 12y = 9

6x = 12y + 9

x = 2y + 3/2 ... Eq.3

Use Eq.3 to substitute to Eq.1 to solve for k

kx - 4y = 3

kx = 4y + 3

k(2y + 3/2) = 4y + 3

k = (4y + 3)/(2y + 3/2)

k = 2


Therefore, the equations are

2x - 4y = 3 and 6x - 12y = 9

Solving for x and y, use substitution method.

Use Eq.3 to substitute to Eq.1

2x - 4y = 3

2x = 4y + 3

2(2y + 3/2) = 4y + 3

4y + 3 = 4y + 3


Yes, and if a consistent system, it has infinite number of solution, it is called dependent. When you graph the equations, both equations represent the same line.

Hope this will be helpful to you