Question

determine the value of k so that the following linear equation have no solution (3k+1)x+3y -2=0 and (k^2+1)+(k-2)y-5=0

Answer 1

Answer:

k= -1

Step-by-step explanation

a1/a2=b1/b2 \neq c1/c2

3k+1/k^2+1=3/k-2

-5k-2=3

k=-1