The number of values of k for which the syst | vedclass

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Question

(b) For infinitely many solutions, the two equations must be identical

$ \Rightarrow \frac{{k + 1}}{k} = \frac{8}{{k + 3}} = \frac{{4k}}{{3k - 1}}$

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(b) For infinitely many solutions, the two equations must be identical

$ \Rightarrow \frac{{k + 1}}{k} = \frac{8}{{k + 3}} = \frac{{4k}}{{3k - 1}}$

$ \Rightarrow (k + 1)(k + 3) = 8k$ and $8\,(3k - 1) = 4k(k + 3)$

$ \Rightarrow {k^2} - 4k + 3 = 0$ and ${k^2} - 3k + 2 = 0$.

By cross multiplication, $\frac{{{k^2}}}{{ - 8 + 9}} = \frac{k}{{3 - 2}} = \frac{1}{{ - 3 + 4}}$

${k^2} = 1$ and $k = 1$;  $k = 1$.

The number of values of $k $ for which the system of equations $(k + 1)x + 8y = 4k,$ $kx + (k + 3)y = 3k - 1$ has infinitely many solutions, is

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