Question

Find the value of k for which the system of equations
kx + 3y = 3, 12x + ky = 6 has no solution.

Answer 1

The given system of equations:
kx + 3y = 3
kx + 3y - 3 = 0                    ….(i)
12x + ky = 6
12x + ky - 6 = 0                    ….(ii)
These equations are of the following 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 and a_2 = 12, b_2 = k, c_2= –6`
In order that the given system has no solution, we must have:
`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`
`i .e.,  k/12 = 3/k ≠ (−3)/(−6)`
`k/12 = 3/k and 3/k ≠ 1/2`
`⇒ k^2 = 36 and k ≠ 6`
`⇒ k = ±6 and k ≠ 6`
Hence, the given system of equations has no solution when k is equal to -6.