Question

For which of following value of k will 3x + (k+3)y = 1 and kx + 6y = 4 have a unique solution?

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Answer 1

Hey friend,

Here is the solution :-

Here, A1 = 3, B1 = k+3, C1 = 1

A2 = k, B2 = 6, C2 = 4

Unique solution = A1/A2 ≠ B1/B2 = C1/C2

Here we take B1/B2 = C1/C2

$\frac{k + 3}{6} = \frac{1}{4} \\ \\ now \: by \: crossmultiplication : \\ \\ 4(k + 3) = 1(6) \\ \\ 4k + 12 = 6 \\ \\ 4k = 6 - 12 \\ \\ 4k = - 6 \\ \\ k = - \frac{6}{4} \\ \\ \\ \\ here \: we \: get \: k = - \frac{6}{4}$

✨I hope this will help you....✨