Find the value of k for which the following pair of equation has no solution. $x + 2 y = 3; (k - 1) x + (k + 1) y k + 2$

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Solution

$x + 2 y = 3$
Comparing with $a_{1} x + b_{1} y + c_{1} = 0$
$a_{1} = 1, b_{1} = 2, c_{1} = 3$
$(k - 1) x + (k + 1) y = k + 2$
Comparing with $a_{2} x + b_{2} y + c_{2} = 0$
$a_{2} = (k - 1), b_{2} = (k + 1), c_{2} = (k + 2)$
To have no solution ,
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \Rightarrow \frac{1}{k - 1} = \frac{2}{k + 1} \Rightarrow k + 1 = 2 k - 2 k = 3.$

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