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Find the value of $k$ for which the following pair of linear equations will have infinitely many solutions :
$kx + 3y - (k - 3) = 0$ and $12x + ky - k = 0$
Hence, find any two solutions of the given pair of equations.
$kx + 3y - (k - 3) = 0$ and $12x + ky - k = 0$
Hence, find any two solutions of the given pair of equations.
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For infinitely many solutions: $\frac{k}{12} = \frac{3}{k} = \frac{k - 3}{k}$ [$1$ mark]
$k^2 = 36$ and $k^2 - 3k = 3k$
$(k = \pm 6)$ and $(k = 6, 0)$
$\therefore k = 6$ [$1$ mark]
For $k = 6$, equations are $6x + 3y = 3$ and $12x + 6y = 6$
any two correct solutions [$\frac{1}{2} + \frac{1}{2}$ mark]
$k^2 = 36$ and $k^2 - 3k = 3k$
$(k = \pm 6)$ and $(k = 6, 0)$
$\therefore k = 6$ [$1$ mark]
For $k = 6$, equations are $6x + 3y = 3$ and $12x + 6y = 6$
any two correct solutions [$\frac{1}{2} + \frac{1}{2}$ mark]