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# 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

 C) 6

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, $$\Large K = \sqrt{36} = 6$$

Part of solved Linear Equations questions and answers : >> Aptitude >> Linear Equations

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