Find the values of ' $k$ ' for which the system of equations $x - 2 y = 3$ , $3 x + k y = 1$ has a unique solution.

k \neq 6

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

k \neq - 6

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

k \neq 5

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

None of these

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

Open in App

Solution

The correct option is B $k \neq - 6$
for unique solution
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \Rightarrow \frac{1}{3} \neq \frac{- 2}{k}$
$\Rightarrow k \neq - 6$

Suggest Corrections

Similar questions

The system x-2y=3 and 3x+ky=1 has a unique solution only when

(a) k=-6 (b) $k \neq - 6$ (c) k=0 (d) $k \neq 0$

k

x - k y = 2

3 x + 2 y = - 5

K

x - K y = 2

3 x + 2 y = - 5

x + 2 y = 5

3 x + k y - 15 = 0

k

x - 2 y = 3

3 x + k y = 1

Join BYJU'S Learning Program