Question

Find the value of k for which the system of linear equations has a unique solution:

(k – 3) x + 3y – k, kx + ky - 12 = 0.

Answer 1

The given system of equations can be written as

(k – 3) x + 3y - k = 0

kx + ky - 12 = 0

This system is of the form:

`"a"_1x+"b"_1"y"+"c"_1 = 0`

`"a"_2x+"b"_2"y"+"c"_2 = 0`

where, `"a"_1 = "k", "b"_1= 3, "c"_1= -"k" and "a"_2 = "k", "b"_2 = "k", "c"_2= -12`

For the given system of equations to have a unique solution, we must have:

`("a"_1)/("a"_2) = ("b"_1)/("b"_2) = ("c"_1)/("c"_2)`

`⇒ ("k"−3)/"k" = 3/"k" = (−"k")/(−12)`

`⇒ "k"  –  3 = 3 and "k"^2 = 36`

⇒ k = 6 and k = ± 6

⇒ k = 6

Hence, k = 6.