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Question

The pair of linear equations $$2x + ky = k, 4x + 2y = k + 1$$ has infinitely many solutions if


A
k=1
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B
k1
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C
k=2
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D
k=4
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Solution

The correct option is B $$k=1$$
The equation are
$$2x+ky-k=0$$
$$4x+2y-(k+1)=0$$
Here, $$ a_{1}=2,b_{1}=k,c_{1}=-k$$
and $$ a _{2}=4,b_{2}=2,c_{2}=-(k+1)$$
For the system to have infinite solutions,
$$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$
$$\Rightarrow \displaystyle \frac{2}{4}=\frac{k}{2}=\frac{-k}{-(k+1)}$$
Taking, 
$$\displaystyle \frac{2}{4}=\frac{k}{2}$$
$$\Rightarrow 4k=4$$
$$\Rightarrow k=1$$   
Taking,
$$\displaystyle\frac{k}{2}=\frac{-k}{-(k+1)}$$
$$\Rightarrow k+1=2 \Rightarrow k=1$$
So, $$k=1$$ is the answer

Mathematics

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