MathematicsThe pair of linear equations 3x+7y=k, 12x+2ky=4k+1 do not have any solution if

The pair of linear equations $3 x + 7 y = k, 12 x + 2 ky = 4 k + 1$ do not have any solution if

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A
$k = 7$
B
$k = 14$
C
$k = 21$
D
$k = 28$

Solution:

The given pair of linear equations are,

3 x + 7 y - k = 0

12 x + 2 ky - (4 k + 1) = 0

Here,

a_{1} = 3, b_{1} = 7, c_{1} = - k

.
And

a_{2} = 12, b_{2} = 2 k, c_{2} = - (4 k + 1)

.
The system has no solution if

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

\Rightarrow \frac{3}{12} = \frac{7}{2 k} \neq \frac{- k}{- (4 k + 1)}

\Rightarrow \frac{1}{4} = \frac{7}{2 k} \neq \frac{k}{4 k + 1}

⇒

2 k = 28

⇒

k = 14

Hence, the correct option is 2.

The pair of linear equations $3 x + 7 y = k, 12 x + 2 ky = 4 k + 1$ do not have any solution if

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