For what value of k the following system of equations has a unique solution $2 x + 3 y - 5 = 0, k x - 6 y - 8 = 0$ ?

k = - 4

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k \neq - 4

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k \neq 4

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k = 4

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Solution

The correct option is B $k \neq - 4$
Given, equations are $2 x + 3 y - 5 = 0$ , $k x - 6 y - 8 = 0$
Here, $a_{1} = 2$ , $b_{1} = 3$ , $c_{1} = - 5$ and $a_{2} = k$ , $b_{2} = - 6$ , $c_{2} = - 8$
For unique solutions, we know
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
$\Rightarrow \frac{2}{k} \neq \frac{3}{- 6}$
$\Rightarrow 3 k \neq - 12$
$\Rightarrow k \neq - 4$

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Similar questions

Find the value for k which each of the following systems of equations has a unique solution:

$2 x + 3 y - 5 = 0, k x - 6 y - 8 = 0.$

2 x + 3 y - 5 = 0

k x - 6 y - 8 = 0

2 x + 3 y - 5 = 0 k x - 6 y - 8 = 0

2 x + 3 y = 4, (k + 2) x + 6 y = 3 k + 2

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