Question

For what value of k does the system of equations $x+2y=5,3x+ky+15=0$ have (i) a unique solution,(ii) no solution?

Answer 1

The equations are written as
x+2y-5=0
3x+ky+15=0

$a_1=1b_1=2c_1=-5$
$a_2=3b_2=k{c}_2=15$

(i) For unique solution,we have

$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
$\frac{1}{3}\ne \frac{2}{k}$

k≠2$\times$3⇒k≠6

Therefore, the given system will have a unique solution for all real values of k other than 6.

(ii) For no solution, we have

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\frac{1}{3}\ne \frac{-5}{15}$
$\frac{1}{3}=\frac{2}{k}$
k=6

Therefore, the given system of equations will have no solutions, if k=6