Find the value of $k$ such that the following pair of linear equations has unique solution.
$4 x + k y + 8 = 0;$ $2 x + 3 y + 7 = 0$

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Solution

Consider the given equations.

$4 x + k y + 8 = 0$ ……. (1)

$2 x + 3 y + 7 = 0$ …… (2)

The general equations,

$a_{1} x + b_{1} y + c_{1} = 0$

$a_{2} x + b_{2} y + c_{2} = 0$

Therefore,

$a_{1} = 4, b_{1} = k, c_{1} = 8$

$a_{2} = 2, b_{2} = 3, c_{2} = 7$

Since, the condition of unique solution is

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

Therefore,

$\frac{4}{2} \neq \frac{k}{3}$

$\frac{k}{3} \neq 2$

$k \neq 6$

Hence, $k$ is any real number except $6$ .

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