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Question

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


A
1
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B
2
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C
3
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D
infinite
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Solution

The correct option is B $$1$$
For no solution
$$\dfrac{k+1}{k}=\dfrac{8}{k+3}\neq\dfrac{4k}{3k-1} \cdots$$ (1)

$$
\Rightarrow \ (k+1)(k+3)-8k=0
$$
or $$k^{2}-4k+3=0= k=1,3$$
But for $$k=1$$, equation (1) is not satisfied Hence $$k=3$$.

Mathematics

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