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प्रश्न
For what values of k will the following pair of linear equations have infinitely many solutions?
kx + 3y – (k – 3) = 0
12x + ky – k = 0
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उत्तर
Here, `a_1/a_2 = k/12, b_1/b_2 = 3/k, c_1/c_2 = (k - 3)/k`
For a pair of linear equations to have infinitely many solutions: `a_1/a_2 = b_1/b_2 = c_1/c_2`
So, we need `k/12 = 3/k = (k - 3)/k`
or `k/12 = 3/k`
Which gives k2 = 36, i.e., k = ± 6.
Also, `3/k = (k - 3)/k`
Gives 3k = k2 – 3k, i.e., 6k = k2, which means k = 0 or k = 6.
Therefore, the value of k, that satisfies both the conditions, is k = 6. For this value, the pair of linear equations has infinitely many solutions.
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