Question

find the value of k for which the equation has no solution kx+3y=3,12x+ky=6​

Answer 1

Given Equation

kx + 3y = 3       (i)

12 + ky = 6        (ii)

To Find the value of k

We Can Write as

kx + 3y -3= 0       (i)

12 + ky-6 = 0        (ii)

For No Solution

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Now Compare with

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

We Get

a₁ = k , b₁ = 3 and c₁= -3

a₂ = 12 , b₂ = k and c₂ = -6

Now Put the value value

a₁/a₂ = b₁/b₂

k/12 = 3/k

k² =36

k = ±6

Answer

k = ±6

Answer 2

Given:

Two sets of equations are given:

• kx+3y = 3
• 12x+ky = 6

To find:

• Value of k for which the equation has no solution.

Solution:

➻ kx+3y = 3

➻ kx+3y-3 = 0

where,

• $\tt{a_1 = k}$
• $\tt{b_1 = 3}$
• $\tt{c_1 = -3}$

Also,

➻ 12x+ky = 6

➻ 12x+ky-6 = 0

where,

• $\tt{a_2 = 12}$
• $\tt{b_2 = k}$
• $\tt{c_2 = -6}$

We know that, if the system of equations has no solution, it is of the form:

$\tt{\longrightarrow \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \cancel{=} \dfrac{c_1}{c_2}}$

Let us substitute the values:

$\tt{\longrightarrow \dfrac{k}{12} = \dfrac{3}{k} \cancel{=} \dfrac{-3}{-6}}$

$\tt{\longrightarrow \dfrac{k}{12} = \dfrac{3}{k}\:and\:\dfrac{3}{k} \cancel{=} \dfrac{-3}{-6}}$

By cross multiplication, we get:

$\tt{\longrightarrow k^2 = 36 \: and \: k \cancel{=} 6}$

$\tt{\longrightarrow k = \pm 6 \: and \: k \cancel{=} 6}$

Hence, the given system of equations has no solution if k = -6.