Question

Find the value of k for which the system of equations
kx + 3y + 3 - k = 0, 12x + ky - k = 0
has no solution.

Answer 1

The given system of equations can be written as
kx + 3y + 3 - k = 0                       ….(i)
12x + ky - k = 0                           ….(ii)
This system of the form:
`a_1x+b_1y+c_1 = 0`
`a_2x+b_2y+c_2 = 0`
where, `a_1 = k, b_1= 3, c_1 = 3 - k and a_2 = 12, b_2 = k, c_2= –k`

For the given system of linear equations to have no solution, we must have:
`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`
`⇒ k/12 = 3/k ≠ (3−k)/(−k)`
`⇒k/12 = 3/k and 3/k ≠ (3−k)/(−k)`
`⇒ k^2 = 36 and -3 ≠ 3 - k`
⇒ k = ±6 and k ≠ 6
⇒k = -6
Hence, k = -6.