Question

For what value of k, the system of equations $x+2y=5,3x+ky+15=0$ has a unique solution?

___

Answer 1

We have,

$x+2y–5=0,3x+ky+15=0.$

The required condition for unique solution is,

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

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

$\Rightarrow k\ne 6$

Hence, for all real values of k except 6, the given system of equations will have a unique solution.