Question

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

$4x+ky+8=0,x+y+1=0.$

Answer 1

The given system may be written as

$4x+ky+8=0,x+y+1=0.$

The given system of equation is of the form

$a_{1}x+b_{1}y+c_1=0,a_{2}x+b_{2}y+c_2=0$

Where,
$a_1=4,b_1=k,c_1=8$

$a_2=1,b_2=1,c_2=1$

For unique solution,we have

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

$\frac{4}{1}\ne \frac{k}{1}$

⇒$k\ne 4$

Therefore, the given system will have unique solution for all real values of k other than 4.