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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.
Solution
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.
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