For what value of k will the system of linear equation has unique solution and no solution x+2y = 5 , 3x+ky =15
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a1=1 b1=2 c1=5
a2=3 b2=k c2=15
⭐If it have unique solution
then, a1/a2≠b1/b2
1/3≠2/k
k≠6
The value of k can be any no. except 6
⭐If it have no solution
then, a1/a2=b1/b2≠c1/c2
firstly taking,
a1/a2=b1/b2
1/3=2/k
k=6
secondly taking,
b1/b2≠c1/c2
2/k≠5/15
2/k≠1/3
k≠6
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