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The number of values of k for which the system of equations $(k+1)x+8y=4k,kx+(k+3)y=3k-1$ has infinitely many solutions is

(A) 0

(B) 1

(C) 2

(D) infinite

1 Answer

For infinitely many solutions the two equation become identical.
$\Rightarrow \large\frac{k+1}{k}=\frac{8}{k+3}=\frac{4k}{3k-1}$
$\Rightarrow k=1$
Hence (b) is the correct answer.
answered Nov 20, 2013 by sreemathi.v

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