Question

for what value of k,will be the following system of equation no solution (3k+1) x+3y--2, (k2+1) x+(k-2) y-5

Answer 1

heya,
it's Aastha......

◆(3k + 1 )x+ 3y-2 ------1
k2+1) x+(k-2) y-5 -------2

◆on comparing the eq 1 nd 2....
a1 = 3k +1 , b1 = 3 , c1 = -2
a2 = k^2 + 1 ,b2 = k-2 , c2 = -5

●the equation have no solution.
so,

◆a1 / a2 = b1 /b2 # c1 / c2

●3k+1 / k ^2 +1 = 3/ k -2 .

◆3k^2 - 6k + k-2 = 3k^2 +3

●-5k -2 = 3

◆-5k = 5

●k = -1

I hope it help you

Answer 2

Answer:

k=-1

Step-by-step explanation:

hope it helps u w8idje8did