Question

Find the valu k such that the equation has no solution (3k+1)x+3y=0 (k²+1)x+(k-2)y

Answer 1

$Heya$‼

$Given\:equations :-$
$(3k+1)x+3y=0$ ...(1)
$(k^2+1)x+(k-2)y=0$ ...(2)

As we know,
conditions for no solution :-
$a1/a2=b1/b2$≠$c1/c2$

Now, $(3k+1)/(k^2+1)=3/(k-2)$≠$0/0$
$(3k+1)/(k^2+1)=3/(k-2)\:and\:3/(k-2)=0$
$(3k+1)/(k^2+1)=3/(k-2)$
$(3k+1)(k-2)=3(k^2+1)$
$3k(k-2)+(k-2)=3k^2+3$
$3k^2-6k+k-2=3k^2+3$
$3k^2-5k-2=3k^2+3$
$3k^2-3k^2-5k-2-3=0$
$-5k-5=0$
$-5k=5$
$k=-5/5$
$k=-1$

Hence, the value of k is –1 .