Question

Find the value of $k$ for which the systerm of equation $x-2y=3$ and $3x+ky=1$ has a unique solution.

Answer 1

We have,

$x-2y+3=0$ and $3x+ky-1=0$

The system of equations will have unique solution

Then, the lines are intersecting.

So,

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$

$\Rightarrow \frac{1}{3}\ne \frac{-2}{k}$

$\Rightarrow k\ne -6$

So, the value of $k$ is $\{k:k\in R,k\ne 6\}$.

Hence, this is the answer.