For what value of k will the following pair of linear equation have no solution?

$2 x + 3 y = 9, 6 x + (k - 2) y = (3 k - 2) .$

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Solution

$2 x + 3 y = 9$

$6 x + (k - 2) y = (3 k - 2)$

Equation can be written as

$2 x + 3 y - 9 = 0$

$6 x + (k - 2) y - (3 k - 2) = 0$

Compare with

$a_{1} x + b y_{1} + c_{1} = 0$

$a_{2} x + b y_{2} + c_{2} = 0$

$a_{1} = 2 b_{1} = 3 c_{1} = - 9$

$a_{2} = 6 b_{2} = k - 2 c_{2} = - (3 k - 2)$

And for no solution

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

Take

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$

$\frac{2}{6} = \frac{3}{k - 2}$

$2 k - 4 = 18$

$2 k = 22$

$k = 11$

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