Question

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

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

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 answerMathematics

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