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
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(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
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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
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