Question

For what value of $k$, the equation $2x-3y=4$ and $3x-ky=5$ meet at exactly one point i.e., point whose abscissa is $-1$?

Answer 1

Given: $x=-1$

Let $2x-3y=4$

$2(-1)-3y=4$

$-2-3y=4$

$-3y=6$

$y=-2$

Now

$3x-yk=5$

$3(-1)-k(-2)=5$

$-3+2k=5$

$2k=8$

$\Rightarrow k=4$