Question

Find the value of k for which the given system has unique solution 2x+3y-5=0 kx-6y-8=0

Answer 1

K ≠ (-4)

Any value of k can be there for unique solution for the given equations except (-4)

So,

One value of k can be 6



SOLUTION :-
$2x + 3y - 5 = 0 \\ \\ kx - 6y - 8 = 0 \\ \\ for \: unique \: solution \: \\ \\ \frac{a1}{a2}\: not \: equal \: to \: \frac{b1}{b2} \\ \\ so \\ \\ \frac{2}{k} \: not \: equal \: to \: \frac{3}{( - 6)} \\ \\ \frac{2}{k} not \: equal \: to \: \frac{( - 1)}{2} \\ \\ so \\ \\k \: not \: equal \: to \: ( - 4)$


Thanks!