System of linear equations

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Question

The number of values of K for which the system of equations $(k + 1) x + 8 y = 4 k, k x + (k + 3) y = 3 k - 1$ has no solution is

Moderate

Solution

From the given system, we have
$\begin{array}{l} \frac{k + 1}{k} = \frac{8}{k + 3} \neq \frac{4 k}{3 k - 1} \\ \Rightarrow k^{2} + 4 k + 3 = 8 k \\ \Rightarrow k = 1, 3 \\ \Rightarrow k = 3 (∴ k = 1 c a n n o t s a t i s t y t h e i n e q u a l i t y) \end{array}$

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