Question

for which value of k will the folowing pair of linear equation have no solution?

1. 3x + y =1 ; (2k-1)x + (k-1)y = 2k +1

2. kx + 3y = 3 ; 12x -6y = 8

Answer 1

$1) \: 3x + y - 1 = 0 \\ \: \: \: \: \: (2k - 1)x + (k - 1)y - (2k + 1) = 0 \\ \\ for \: no \: solution \\ = > \frac{a1}{a2} = \frac{b1}{b2} \: ≠ \: \frac{c1}{c2} \\ = > \frac{3}{2k - 1} = \frac{1}{k - 1} \: not equal to \: \frac{ - 1}{ - (2k + 1)} \\ = > \frac{3}{2k - 1} = \frac{1}{k - 1} \\ by \: cross \: multiplication \\ = > 3(k - 1) = (2k - 1) \\ = > 3k - 3 = 2k - 1 \\ = > 3k - 2k = - 1 + 3 \\ = > k = 2$
$2)kx + 3y - 3 = 0 \\ \: \: \: 12x - 6y - 8 = 0 \\ for \: no \: solution \\ = > \frac{a1}{a2} = \frac{b1}{b2} \: ≠ \: \frac{c1}{c2} \\ = > \frac{k}{12} = \frac{3}{ - 6} \: ≠ \: to \: \frac{ - 3}{ - 8} \\ = > \frac{k}{12} = \frac{3}{ - 6} \\ by \: cross \: multiplication \\ = > - 6k = 36\\ = >k = \frac{36}{ - 6} \\ = > k = - 6$