Question

for what value of k will the following system of equations have no solutions (3k+1)x +3y=2 and (k^2+1)x +(k-2)y=5​

Answer 1

Answer:

(3k+1)x+3y-2=0 and (k²+1)x+(k-2)y-5 =0 ... 1 answer. For what value of K will the following pair of linear equation have no

Answer 2

Answer:

The condition for having no solution: a1/a2 = b1/b2 ≠ c1/c2

equations= (3k+1)x + 3y - 2=0   and   (k∧2+1)x +(k-2)y - 5=0

a1= (3k+1)           a2= (k∧2+1)

b1= 3                 b2= (k-2)

c1= -2                c2= -5

step 1: a1/a2 = b1/b2

          (3k+1)/ (k∧2+1)  = 3/(k-2)

          (3k-1)×(k-2)        =3×(k∧2+1)

          3k∧2 -6k +k -2 = 3k∧2 + 3

          3k∧2 -3k∧2 -5k= 3+2

          0k∧2 -5k           = 5

                              k    =5/-5 = -1

value of k=-1