The value of k for which $k x + 3 y - k + 3 = 0$ and $12 x + k y = k$ , have infinite solutions, is?

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $6$
For infinite many solution $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$\Rightarrow \frac{k}{12} = \frac{3}{k} = \frac{- (k - 3)}{- k}, \frac{3}{k} = \frac{(k - 3)}{k}$
$k^{2} = 36$ $k - 6 = 0$
$k = \pm 6$ $k = 6$ .

Suggest Corrections

Similar questions

K

K x + 3 y - (K - 3) = 0

12 x + K y - K = 0

k

k x + 3 y = k - 3

12 x + k y = k

k x + 3 y - (k - 3) = 0, 12 x + k y - k =

k =

k x + 3 y = k - 3; 12 x + k y = k; k \neq 0

k x + 3 y = k - 3

12 x + k y = k

Join BYJU'S Learning Program