Find the value of k for …
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Sia ? 6 years, 6 months ago
The given pair of linear equation is
3x + y = 1, (2k - 1) x + (k - 1) y = 2k + 1
{tex}\Rightarrow{/tex} 3x + y - 1 = 0
Here, (2k - 1)x + (k - 1) y - (2k + 1) = 0
a1 = 3, b1 = 1, c1 = -1
a2 = 2k - 1, b2 = k - 1, c2 = -(2k + 1)
for having no solution, we must have {tex}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}{/tex}
{tex}\Rightarrow \frac{3}{{2k - 1}} = \frac{1}{{k - 1}} \ne \frac{{ - 1}}{{ - (2k + 1)}}{/tex}
From above we have {tex}\frac{3}{{2k - 1}} = \frac{1}{{k - 1}}{/tex}
{tex}\Rightarrow{/tex} 3(k - 1) = 2k - 1
{tex}\Rightarrow{/tex} 3k - 3 = 2k - 1
{tex}\Rightarrow{/tex} 3k - 2k = 3 - 1
{tex}\Rightarrow{/tex} k = 2
Hence, the required value of k is 2.
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