Question

kx+3y=k-3 and 12x+ky=6​

Answer 1

Answer:

x=1/2-1/12ky

Step-by-step explanation:

Answer 2

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Step-by-step explanation:

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: ky + 3y = 3$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 12x+ ky \: = 6$

$Here, \: \: \: a_{1} = k, \: b_{1} = 3, \: \: c_{1} = 3 \\ a_{2} = 12, \: b_{2} = k, \: c_{2} = 6$

∵ Equation has no solution,

$∴ \: \: \: \: \frac{ a_{1} }{ a_{2}}= \frac{ b_{1} }{ b_{2} } ≠ \frac{ c_{1}}{ c_{2} }$

$⟹ \: \: \: \frac{k}{12} = \frac{3}{k} ≠ \frac{3}{6}$

$⟹ \: \: \: \frac{k}{12} = \frac{3}{k}$

$⟹ \: \: \: \: {k}^{2} = 36$

$∴ \: \: \: \: \: k = ±6 \: \: \: \: \: \: \: \: \: \: ....(1)$

$and \: \: \: \: \: \: \: \: \frac{3}{k} ≠ \frac{3}{6}$

$⟹ \: \: \: \: \frac{1}{k} ≠ \frac{1}{6}$

$⟹ \: \: \: \: \: k≠6 \: \: \: \: \: \: \: \: \: \: \: ....(2)$

Therefore from eqns. (1) and (2)

$\: \: \: \: \: \: \: \: \: \: \: \: k = - 6.$