TS ICET 24 May 2019 Shift 1 Paper

Question : 126 of 200

Marks: +1, -0

The value of k for which the system kx + 3y = 3 and 12x + ky = k of equations has infinitely many solutions is:

Solution:

Given equations are

k x+3 y-(k-3)=0

12 x+k y-k=0

Since above equations have infinite solutions

\;{k}/{12}=\;{3}/{k}=\;{-(k-3)}/{-k}

k ^{2}=36 ⇒ k =± 6

→ (1)

\;{3}/{k}=\;{(k-3)}/{k} ⇒ 3=k-3

k=6

→ (2)
From (1) & (2), k = 6.

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