Aspire Faculty ID #3626 · Topic: NIMCET 2012 · Just now
NIMCET 2012

The number of values of $k$ for which the system of equations $(k+1)x + 8y = 4k$ and $kx + (k+3)y = 3k-1$ has infinitely many solutions, is

Solution

$ (k+1)x + 8y = 4k $ 
$ kx + (k+3)y = 3k - 1 $ 
Since the system has infinitely many solutions, $ \dfrac{k+1}{k} = \dfrac{8}{k+3} = \dfrac{4k}{3k-1} $ 
Taking 1st and 3rd ratio: $ (k+1)(3k-1) = 4k^2 $ 
$ 3k^2 + 2k - 1 = 4k^2 $ 
$ k^2 - 2k + 1 = 0 $ 
$ (k-1)^2 = 0 $ 
$ \therefore k = 1 $

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