Question

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

kx + 3y – (k – 3) = 0

12x + ky – k = 0

Answer 1

Here, `a_1/a_2 = k/12, b_1/b_2 = 3/k, c_1/c_2 = (k - 3)/k`

For a pair of linear equations to have infinitely many solutions: `a_1/a_2 = b_1/b_2 = c_1/c_2`

So, we need `k/12 = 3/k = (k - 3)/k`

or `k/12 = 3/k`

Which gives k² = 36, i.e., k = ± 6.

Also, `3/k = (k - 3)/k`

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.