For what value of k, will the following system of equations 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
Condition for infinitely many solutions is

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

$2 x + 3 y = 4 a n d (k + 2) x + 6 y = 3 k + 2$

$\frac{2}{k + 2} = \frac{4}{3 k + 2}$

$\Rightarrow 6 k + 4 = 4 k + 8$

$\Rightarrow$ $k = 2$

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Similar questions

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

2 x + 3 y = 4

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

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

Find the value of k for which the following pair of linear equations have infinitely many solutions:

$2 x + 3 y = 7, (k - 1) x + (k + 2) y = 3 k .$

k

2 x + 3 y = 7, (k - 1) x + (k + 2) y = 3 k

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