Question

find the value of K for which the pair of linear equations
kx+3y=k-2 and 12x+ky=k
has no solution.​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

The given equation are

kx+3y+(2-k)=0

12x+ky-k=0

These equation are of the form

$\sf \bullet a_1x+b_1y+c_1=0 \: \: and \: \: a_2x+b_2y+c_2=0$

Where,

$\sf a_1=k,a_2=12 \\ \\ \sf b_1=3,b_2=k \\ \\ \sf c_1=(2-k),c_2=-k$

In order that the given system has no solution, we must have

$\longrightarrow \sf \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2}$

This happens when

$\sf \dfrac{k}{12} = \dfrac{3}{k} \ne \dfrac{(k - 2)}{k}$

$\sf \: when \: \dfrac{k}{12} = \dfrac{3}{k} \: and \: \dfrac{3}{k} \ne \dfrac{(k - 2)}{k}$

$\sf \: but \: \: \dfrac{k}{12} = \dfrac{3}{k} \implies {k}^{2} = 36 \implies \: k = \pm6$

Case 1- When k=6,we have;

$\: \: \sf \dfrac{3}{k} = \dfrac{3}{6} = \dfrac{1}{2} \: and \: \dfrac{(k - 2)}{k} = \dfrac{4}{6} = \dfrac{2}{3}$

$\therefore \sf \dfrac{k}{12} = \dfrac{3}{k} \ne \dfrac{(k - 2)}{k}$

Case-2- When k=-6, we have;

$\: \: \sf \dfrac{3}{k} = \dfrac{3}{ - 6} = \dfrac{ - 1}{2} \: and \: \dfrac{(k - 2)}{k} = \dfrac{ - 8}{ - 6} = \dfrac{4}{3}$

$\therefore \sf \dfrac{k}{12} = \dfrac{3}{k} \ne \dfrac{(k - 2)}{k}$

Thus, in each case ,the given system has no solution

Hence, the required value of k are 6 and -6