The value of k for which the system of equations 3x – y + 8 = 0 and 6x – ky + 10 = 0 has infinitely many solutions, is

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Solution

Q.
The value of k for which the system of equations 3x – y + 8 = 0 and 6x – ky + 10 = 0 has infinitely many solutions, is

1 – 2

2 2

3 1/2

4 − 1/2

Ans.
(2)

Given equation are $3 x - y + 8 = 0$ and $6 x - k y + 16 = 0$

Here, $a_{1} = 3, b_{1} = - 1, c_{1} = 8$

$a_{2} = 6, b_{2} = - k, c_{2} = 16$

For Infinite many solution, $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \Rightarrow \frac{3}{6} = \frac{- 1}{- k} = \frac{8}{16} \Rightarrow \frac{1}{k} = \frac{1}{2} \Rightarrow k = 2$

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