Question

For what values of K will the following pair of linear equations have infinitely many solutions?

kx + 3y – (k – 3) = 0 and 12x + ky – k = 0​

Answer 1

$\implies$$here \: \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}$

$\implies$ For a pair of linear equations to have infinitely many solution

So , we need $\frac{k}{12} = \frac{3}{k} \frac{k - 3}{k}$

or ,

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

$\implies$ Which gives

$k {}^{2} = 36 \: ie \: k = \frac{ - }{ + } 6$

$\implies$ Also

$\frac{3}{k} \: \: = \: \frac{k - 3}{k}$

$\implies$ gives 3k = k2 – 3k, i.e., 6k = k2 , which means k = 0 or k = 6.

Therefore, the value of k, that satisfies both the conditions, is k = 6. For this value, the pair of linear equations has infinitely many solutions.