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Question
Find the value of k for which the system of equations has a unique solution:
4x + ky + 8=0,
x + y + 1 = 0.
Solution
The given system of equations are
4x + ky + 8 = 0
x + y + 1 = 0
This system is of the form:
`a_1x+b_1y+c_1 = 0`
`a_2x+b_2y+c_2 = 0`
where, `a_1 = 4, b_1= k, c_1 = 8 and a_2 = 1, b_2 = 1, c_2 = 1`
For the given system of equations to have a unique solution, we must have:
`(a_1)/(a_2)≠ (b_1)/(b_2)`
⇒`4/1≠k/1`
⇒ k ≠ 4
Hence, k ≠ 4.
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