Question

For the following system of equation determine the value of k for which the given system of equation has a unique solution.
$2x+3y-5=0$
$kx-6y-8=0$

Answer 1

The given system of equations is
$2x+3y-5=0$
$kx-6y-8=0$

It is of the form $a_{1}x+b_{1}y+c_1=0$
and, $a_{2}x+b_{2}y+c_2=0$

where, $a_1=2,b_1=3,c_1=-5$

and $a_2=k,b_2=-6$ and $c_2=-8$

For a unique solution, we must have
$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$ i.e., $\frac{2}{k}\ne \frac{3}{-6}\Rightarrow k\ne -4$

So, the given system of equations will have a unique solution for all real values of $k$ other than $-4$.