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Question

Find the value of k for which each of the following system of equations has no solution:
3x+y=1, (2k-1)x+(k-1)y=(2k+1).

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Solution

The given system of equations:
3x + y = 1
⇒ 3x + y − 1 = 0 ...(i)
And, (2k − 1)x + (k − 1)y = (2k + 1)
⇒ (2k − 1)x + (k − 1)y − (2k + 1) = 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = 3, b1= 1, c1 = −1 and a2 = (2k − 1), b2 = (k − 1), c2 = −(2k + 1)
In order that the given system of equations has no solution, we must have:
a1a2=b1b2c1c2
i.e. 32k-1=1k-1-1-2k+1
32k-1=1k-1 and 1k-1-1-2k+1
⇒ 3(k − 1) = 2k − 1 and −(2k + 1) ≠ −(k − 1)
⇒ 3k − 3 = 2k − 1 and −2k − 1 ≠ −k + 1
⇒ k = 2 and k ≠ −2
Thus, a1a2=b1b2c1c2 holds when k is equal to 2.
Hence, the given system of equations has no solution when the value of k is 2.

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