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Question
Find the value of k for which the system of equations has a unique solution:
kx + 3y = (k – 3),
12x + ky = k
Solution
The given system of equations:
kx + 3y = (k – 3)
⇒ kx + 3y – (k - 3) = 0 ….(i)
And, 12x + ky = k
⇒12x + ky - k = 0 …(ii)
These equations are of the following form:
`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`
Here, `a_1 = k, b_1= 3, c_1= -(k – 3) and a_2 = 12, b_2 = k, c_2= -k`
For a unique solution, we must have:
`(a_1)/(a_2) ≠ (b_1)/(b_2)`
i.e., `k /12 ≠ 3/k`
⇒ `k^2 ≠ 36 ⇒ k ≠ ±6`
Thus, for all real values of k, other than ±6, the given system of equations will have a unique solution.
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