Question

Find the value of k for which the given system of equation infinately many solution Kx+3y=k-3
12x+ky=k

Answer 1

Answer:

a1/a2=b1/b2=c1/c2

K/12=3/k=k-3/k

k/12=3/k

k^2=36

k=6

Please mark it as a brainliests answer

Answer 2

Given :

The given system of equations is

kx + 3y + (3 - k) = 0, 12x + ky - k = 0

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Solution:

These equations are of the form

$\sf \: a_1 + b_1y + c_1 = 0 \: and \\ \sf \: a_2 x+ b_2y + c_2 = 0$

Where $\sf a_1 = k, b_1 = 3, c_1 = (3 - k)$

$\sf and a_2 = 12, b_2 = k, c_2 = -k.$

Let the given system of equations have infinity many solutions.

Then,

$\sf \: \frac{a_1 }{a_2 } = \frac{b_1 }{b_2 } = \frac{c_1 }{c_2 }$

$→ \bf\frac{k}{12} = \frac{3}{k} = \frac{(3 - k)}{ - k}$

$→ \bf \frac{k}{12} = \frac{3}{k} = \frac{k - 3}{k}$

$→ \bf{ \frac{k}{12} = \frac{3}{k}} \: \sf{and}\: \bf{ \frac{3}{k } = \frac{k - 3}{k} }$

$→ \bf {{k}^{2} = 36} \: \sf{ and} \: \bf{ {k}^{2} - 6k = 0 }$

$→\bf {(k = 6 \: or \: k = - 6)} \: \\ \sf{and} \: \bf{k(k - 6) = 0}$

$→ \bf{(k = 6 \: or \: k = - 6)} \\ \sf{and} \: \bf{k = 0 \: or \: k = 6}$

$\fbox{→ \: \bf \: k = 6}$

Hence, the given system of equations has infinitely many solutions when k = 6.

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Answer:

The value of k = 6.