Question

find the value of k so that the equation has no solution (3k+1)x+3y-2=0 ;( k square+1)x+(k-2)y-5=0

Answer 1

$Heya$‼

$Given\:equations :-$
$(3k+1)x+3y-2=0$ ...(1)
$(k^2+1)x+(k-2)y-5=0$ ...(2)

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

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

Hence, the value of k is –1 .