For what value of k will the pair of linear equations have infinitely many solution : (K-3)x + 3y = k Kx + KY = 12
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For infinitely many solution
Ratio of a¹/a²=b¹/b²=c¹/c²
Then
(k-3)/k=3/k=k/12
(k-3)/k=3/k
Cross multiplying
k(k-3)=3k
cancel k both sides
k-3=3
k=6
Hope you like it....
Ratio of a¹/a²=b¹/b²=c¹/c²
Then
(k-3)/k=3/k=k/12
(k-3)/k=3/k
Cross multiplying
k(k-3)=3k
cancel k both sides
k-3=3
k=6
Hope you like it....
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