Question

Find k for which the system kx-y=2 and 6x-2y=3 has a unique solution.

Answer 1

Given system of equations are

$6x-2y=3$

$6x-2y-3=0$ ----( 1 )

$kx-y=2$

$kx-y-2=0$ ----( 2 )

Compare above equations with

$a1x+b1y+c1=0$ and

$a2x+b2y+c2=0$ , we get

$a1=6,b1=-2,c1=-3$ ;

$a2=k,b2=-1,c2=-2$ ;

Now ,

$\frac{a1}{a2}\ne \frac{b1}{b2}$

[ Given they have Unique solution ]

$\frac{6}{k}\ne \frac{-2}{-1}$

$\frac{6}{k}\ne 2$

$\frac{k}{6}\ne \frac{1}{2}$

k ≠ 6/2

k ≠ 3

Therefore,

For all real values of k , except k≠ 3,

Above equations have a unique solution.