The value of k for which kx+ 3y-k+ 3=0 and 12x+ky=k, have infinite solutions, is
A) 0
B) -6
C) 6
D) 1
Correct Answer:
C) 6
Description for Correct answer
For infinite solution
\( \Large \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \)
= \( \Large \frac{K}{12}=\frac{3}{K}=\frac{-K+3}{-K} \)
= \( \Large \frac{K}{12}=\frac{3}{K}=K^{2}=36 \)
Therefore, \( \Large K = \sqrt{36} = 6 \)
Please provide the error details in above question
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