Find the value of k for which the given system of equations has no solution?
$x + 2 y = 3$
$3 x + k y - 15 = 0$

k \neq 6

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k \neq 7

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k = 6

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None of the above

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Solution

The correct option is C $k = 6$
Give: $x + 2 y = 3$
$3 x + k y - 15 = 0$
Condition for no solution:
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

$a_{1} = 1, a_{2} = 3, b_{1} = 2, b_{2} = k$

$\frac{1}{3} = \frac{2}{k} \neq \frac{3}{15} \Rightarrow k = 6$

When $k = 6$ the set of equations represents two parallel lines which don't intersect at any point. So they can't have a common point which would be the solution.

Suggest Corrections

Similar questions

x + 2 y = 3

3 x + k y - 15 = 0

x + 2 y = 5

3 x + k y - 15 = 0

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

a) unique solution

b) no solution

k

x - k y = 2

3 x + 2 y = - 5

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

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