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

0

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1

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2

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i n f i n i t e

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Solution

The correct option is D $1$
For infinitely many solutions,
$\frac{k + 1}{k} = \frac{8}{k + 3} = \frac{4}{3 k - 1}$
$\Rightarrow k^{2} - 4 k + 3 = 0$ and $24 k - 8 = 4 k + 12$
$\Rightarrow k = 3, k = 1$ and $k = 1$
Hence, $k = 1$

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(k + 1) x + 8 y = 4 k

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\Rightarrow

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