The given pair of linear equation is3x + y = 1, (2k – 1) x + (k – 1) y = 2k + 1{tex}\\Rightarrow{/tex} 3x + y – 1 = 0Here, (2k – 1)x + (k – 1) y – (2k + 1) = 0a1 = 3, b1 = 1, c1 = -1a2 = 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\xa0{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 = 2Hence, the required value of k is 2.

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