Question

find the value of k for which the pair of linear equations 3x + (k + 1)y = 5k + 1 , (2k+1)x + 15y = 63 have infinitely many solutions

Answer 1

Two linear equations of form
a₁x + b₁y + c₁ = 0a₂x + b₂y + c₂ = 0
have infinite solutions if
a₁/a₂ = b₁/b₂ = c₁/c₂
So, putting equation in required form we get
3x + (k+1)y - 5k - 1 = 0(2k+1)x + 15y - 63 = 0
So, we have
3/(2k+1) = (k+1)/15 = -5k - 1/-63
Taking 2nd and 3rd parts we get
15(-5k - 1) = -63(k+1)
-75k - 15 = -63k - 63
-12k = -48
k = 4

Answer 2

i hope it is the right answer