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For which value(s) of k will the pair of equations $kx + 3 y = k– 3; 12 x + ky = k$ have no solution?

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-6

answer is B.

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Detailed Solution

Given pair of linear equations are,

kx + 3 y = k– 3

… (1)

12 x + ky = k

… (2)
Given pair of linear equations have no solution.
On comparing the given pair of linear equations with

ax + by + c = 0

we get,

a_{1} = k, b_{1} = 3, c_{1} = - (k - 3)

a_{2} = 12, b_{2} = k, c_{2} = - k

Then,

\frac{a_{1}}{a_{2}} = \frac{k}{12}

\frac{b_{1}}{b_{2}} = \frac{3}{k}

\frac{c_{1}}{c_{2}} = \frac{k - 3}{k}

We know that for no solution of the pair of linear equations,

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

⇒ k 12 = 3 k ≠ k − 3 k

Taking the second and third ratios,

⇒ 3 k ≠ k − 3 k

⇒ 3 k ≠ k − 3 k ⇒ k 2 − 3 k ≠ 3 k ⇒ k 2 − 3 k − 3 k ≠ 0

⇒ k 2 − 6 k ≠ 0 ⇒ k k − 6 ≠ 0 ⇒ k ≠ 0, 6

Now, taking the first and second ratios,

\frac{k}{12} = \frac{3}{k}

\Rightarrow k^{2} = 36

\Rightarrow k = \pm 6

So, the common value of

k

for which the system of linear equations has no solutions is

k = - 6

.
Hence, the correct option is 2.

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