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Question

Find the value of k for which each of the following system of linear equations has an infinite number of solutions:
(k-1)x-y=5,(k+1)x+(1-k)y=(3k+1).

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Solution

The given system of equations:
(k − 1)x − y = 5
⇒ (k − 1)x − y − 5 = 0 ...(i)
And, (k + 1)x + (1 − k)y = (3k + 1)
⇒ (k + 1)x + (1 − k)y − (3k + 1) = 0 ...(ii)
These equations are of the following form:
a1x + b1y + c1 = 0, a2x + b2y + c2 = 0
Here, a1 = (k − 1), b1= −1, c1 = −5 and a2 = (k + 1), b2 = (1 − k), c2 = −(3k + 1)
For an infinite number of solutions, we must have:
a1a2=b1b2=c1c2
i.e. k-1k+1=-1-k-1=-5-3k+1
k-1k+1=1k-1=53k+1

Now, we have the following three cases:
Case I:
k-1k+1=1k-1
k-12=k+1
k2+1-2k=k+1
k2-3k=0k(k-3)=0k=0 or k=3

Case II:
1k-1=53k+1
3k+1=5k-1
3k+1=5k-5
2k=6k=3

Case III:
k-1k+1=53k+1
3k+1k-1=5k+1
3k2+k-3k-1=5k+5
3k2-2k-5k-1-5=0
3k2-7k-6=0
3k2-9k+2k-6=0
3kk-3+2k-3=0
k-33k+2=0
k-3=0 or 3k+2=0
k=3 or k=-23

Hence, the given system of equations has an infinite number of solutions when k is equal to 3.

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