Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

$k x + 3 y = (2 k + 1) 2 (k + 1) x + 9 y = (7 k + 1) .$

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Solution

The given system may be written as

$k x + 3 y - 2 k + 1 = 0 2 (k + 1) x + 9 y - (7 k + 1) = 0$

The given system of equation is of the form

$a_{1} x + b_{1} y - c_{1} = 0 a_{2} x + b_{2} y - c_{2} = 0$

Where,
$a_{1} = k, b_{1} = 3, c_{1} = - (2 k + 1) a_{2} = 2 (k + 1), b_{2} = 9, c_{2} = - (7 k + 1)$

For unique solution,we have

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \frac{k}{2 (k + 2)} = \frac{3}{9} = \frac{- (2 k + 1)}{- (7 k + 1)} \frac{k}{2 (k + 2)} = \frac{3}{9} a n d \frac{3}{9} = \frac{(2 k + 1)}{(7 k + 1)} \Rightarrow 9 k = 3 \times 2 (k + 1) a n d 3 (7 k + 1) = 9 (2 k + 1) \Rightarrow 9 k - 6 k = 6 a n d 21 k - 18 k = 9 - 3 \Rightarrow 3 k = 6 a n d 3 k = 6 \Rightarrow k = 2 a n d k = 2$

Therefore, the given system of equations will have infinitely many solutions, if k = 2

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Similar questions

Find value of k for infinite Sol for pair of linear equation

kx + 3y = 2k + 1

2 ( k + 1)x + 9x = 7k + 1

x + (k + 1) y = 5

(k + 1) x + 9 y = 8 k - 1

(k - 1) x - y = 5, (k + 1) x + (1 - k) y = (3 k + 1) .

(a - 1) x + 3 y = 2, 6 x + (1 - 2 b) y = 6 .

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