Find the value of k for which following system of equations have infinitely many solutions
$2 x + 3 y - 5 = 0$
$6 x + k y - 15 = 0$

k = 3

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k = 9

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k = 4

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k = 2

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Solution

The correct option is B k = 9
Given: $2 x + 3 y - 5 = 0$
$6 x + k y - 15 = 0$

Condition for infinitely many solutions is:

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

$a_{1} = 2, a_{2} = 6, b_{1} = 3, b_{2} = k$

$\frac{2}{6} = \frac{3}{k} \Rightarrow k = 9$

When $k = 9$ , it represents two exactly same lines, so all the points coincide, thus giving infinitely many solutions.

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2 x + 3 y - 5 = 0

6 x + k y - 15 = 0

2 x + 3 y - 5 = 0

6 x + k y - 15 = 0

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Find the value of k for which each of the following systems of equations have infinitely many solution: (9-19)

2x + 3y - 5 =0

6x + ky - 15 = 0

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k x + k y = 12