For what value of k will the following pair of linear equations have no solution?:
$2 x + 3 y = 9, 6 x + (k - 2) y = (3 k - 2)$

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Solution

The given pair of linear equations are
$2 x + 3 y - 9 = 0 . . . . . (i) 6 x + (k - 2) y - (3 k - 2) = 0 . . . . . (i i)$
Which is of the form $a_{1} x + b_{1} y + c_{1} = 0 and a_{2} x + b_{2} y + c_{2} = 0$ , where $a_{1} = 2, b_{1} = 3, c_{1} = - 9, a_{2} = 6, b_{2} = k - 2 and c_{2} = - (3 k - 2)$
For the given pair of linear equations to have no solution, we must have
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \Rightarrow \frac{2}{6} = \frac{3}{k - 2} \neq \frac{- 9}{- (3 k - 2)} \Rightarrow \frac{2}{6} = \frac{3}{k - 2}, \frac{3}{k - 2} \neq \frac{- 9}{- (3 k - 2)} \Rightarrow k = 11, \frac{3}{k - 2} \neq \frac{9}{3 k - 2}$
$\Rightarrow k = 11, 3 (3 k - 2) \neq 9 (k - 2) \Rightarrow k = 11, 1 \neq 3 (true)$
Hence, k = 11.

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