Question

Find the value of k for which the following pair of linear equations will have infinitely many solutions: $k x + 3 y - (k - 3) = 0$ and $12 x + k y - k = 0$ . Hence, find any two solutions of the given pair of equations.

Approach

For infinitely many solutions, $\frac{k}{12} = \frac{3}{k} = \frac{k - 3}{k}$ . From $\frac{k}{12} = \frac{3}{k}$ we get $k^{2} = 36 \Rightarrow k = \pm 6$ . From $\frac{3}{k} = \frac{k - 3}{k}$ we get $3 = k - 3 \Rightarrow k = 6$ (only positive value satisfies). For k = 6, equations reduce to $6 x + 3 y = 3$ and $12 x + 6 y = 6$ (same line). Two solutions: (0, 1) and (1, -1).

Step-by-step working

1
For infinitely many solutions: $\frac{k}{12} = \frac{3}{k} = \frac{k - 3}{k}$
0.5 marks
2
$k^{2} = 36$ and $k^{2} - 6 k = 0 \Rightarrow k = 6$
1 mark
3
For k = 6, equations: $6 x + 3 y = 3$ and $12 x + 6 y = 6$
0.5 marks
4
Any two correct solutions
1 mark

Concepts used

linear equationsinfinitely many solutionscondition

Approach

Step-by-step working

Concepts used

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