For what value of k, the system of equations x + 2y = 5, 3x + ky + 15 = 0 has no solution?

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Solution

We have,

x + 2 y – 5 = 0 and 3 x + k y + 15 = 0.

The required condition for no solution is $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$\frac{1}{3} = \frac{2}{k} \neq - \frac{5}{15}$

k = 6 & $\frac{2}{k} \neq - \frac{1}{3}$

k = 6 and k ≠ –6

Hence the given system of equations will have no solution when k = 6.

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For what value of k, the system of equations $x + 2 y = 5, 3 x + k y + 15 = 0$ has a unique solution?

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$x + 2 y = 3$
$3 x + k y - 15 = 0$

x + 2 y = 3

3 x + k y - 15 = 0

k

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