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Question

The number of values of k for which the system of equations (k+1)x+8y=4k,kx+(k+3)y=3k1 has no solution is

A
1
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B
3
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C
2
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D
infinite
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Solution

The correct option is A 1
Let =k+18kk+3
=k2+4k+3=(k1)(k3)

For the system of equation to have no solution
=0k=3,1

For k=1, the system of equation becomes
2x+8y=4 and x+4y=2
x+4y=2,x+4y=2
This system has infinite number of solution

For k=3, the system of equation becomes
4x+8y=12 and 3x+6y=8
x+2y=3,x+2y=83
Thus, in this case the system of equation has no solution.

The number of values of k for which the given system of equations has no solution is 1.

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