Question

For the following system of equation determine the value of $k$ for which the given system of the equation has a unique solution.
$x-ky=2$
$3x+2y=-5$

Answer 1

The given system of equations is
$x-ky-2=0$
$3x+2y+5=0$

This system of equation is of the form
$a_{1}x+b_y+c_1=0$
$a_{2}x+b_{2}y+c_2=0$

where, $a_1=1,b_1=-k,c_1=-2$ and $a_2=3,b_2=2,c_2=5$

For a unique solution, we must have

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$ i.e., $\frac{1}{3}\ne \frac{-k}{2}\Rightarrow k\ne \frac{-2}{3}$

So, the given system of equations will have a unique solution for all real values of $k$ other than $-2/3$.