For what value of k, will the following pair of linear equations in two variable have infinitely many solutions?

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

k = 2

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k = 3

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k = 4

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k = 5

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Solution

The correct option is A k = 2
Given: $2 x + 3 y = 4, (k + 2) x + 6 y = 3 k + 2$

Condition for infinitely many solutions is

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

$a_{1} = 2, a_{2} = k + 2, b_{1} = 3, b_{2} = 6, c_{1} = - 4, c_{2} = - (3 k + 2)$

$\frac{2}{k + 2} = \frac{3}{6} = \frac{- 4}{- (3 k + 2)}$

$\frac{2}{k + 2} = \frac{3}{6}$

$\frac{2}{k + 2} = \frac{1}{2}$

$k + 2 = 4$

$k = 2$

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