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# (ii) For which value of k will the following pair of linear equations have no solution? 3x + y = 1 (2k - 1) x + (k - 1) y = 2k + 1

Q2.    (ii) For which value of k will the following pair of linear equations have no solution?

$\\3x + y = 1 \\(2k - 1) x + (k - 1) y = 2k + 1$

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Given, the equations,

$\\3x + y = 1 \\(2k - 1) x + (k - 1) y = 2k + 1$

As we know, the condition for equations $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$  to have no solution is

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

So, Comparing these equations with, $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$, we get

$\frac{3}{2k-1}=\frac{1}{k-1}\neq\frac{1}{2k+1}$

From here we get,

$\frac{3}{2k-1}=\frac{1}{k-1}$

$\Rightarrow 3(k-1)=2k-1$

$\Rightarrow 3k-3=2k-1$

$\Rightarrow 3k-2k=3-1$

$\Rightarrow k=2$

Hence, the value of K is 2.

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