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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

Answers (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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