NIMCET 2012 Question Paper with solutions

Question : 2 of 120

Marks: +1, -0

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

Solution:

Since, the given system has infinitely many solutions
∴

{k+1}/k

8/{k+3}

{4k}/{3k-1}

Taking 1st and IIIrd part,

(k + 1) (3k - 1) =

4k^2

⇒

3k^2

+ 2k - 1 =

4k^2

⇒

k^2

- 2k + 1 = 0
⇒

(k-1)^2

= 0
∴ k = 1

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