Question

Find the value for k which each of the following systems of equations has a unique solution:

$2x+3y-5=0,kx-6y-8=0.$

Answer 1

$2x+3y-5=0$......(1)
$kx-6y-8=0$......(2)

$a_1=2,b_1=3,c_1=-5$
$a_2=k,b_2=-6,c_2=-8$

For a unique solution, we must have

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}or\frac{2}{k}\ne \frac{3}{-6}$

$k\ne \frac{-12}{3}$

$k\ne -4$

This happens when,

$k\ne -4$

Thus, for all real value of k other than the given system of equations will have a unique solution.