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

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

A
$k = 1$
B
$k \neq 1$
C
$k = 2$
D
$k = 4$

Solution:

The given equation are

2 x + ky - k = 0

4 x + 2 y - (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,

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

Here,

\frac{2}{4} = \frac{k}{2} = \frac{- k}{- (k + 1)}

Taking,

\frac{2}{4} = \frac{k}{2}

⇒

4 k = 4

⇒

k = 1

Taking,

\frac{k}{2} = \frac{- k}{- (k + 1)}

⇒

k + 1 = 2

⇒

k = 1

Hence, the correct option is 1.

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