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
B
2
C
3
D
infinite

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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