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

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

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Solution

The given system may be written as

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

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, $a_{1} = 2, b_{1} = 3, c_{1} = - k a_{2} = k - 1, b_{2} = k + 2, c_{2} = - 3 k$

For unique solution,we have

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \frac{2}{k - 1} = \frac{3}{k + 2} = \frac{- k}{- 3 k} \frac{2}{k - 1} = \frac{3}{k + 2} a n d \frac{3}{k + 2} = \frac{k}{3 k} 2 (k + 2) = 3 (k - 1) a n d 3 \times 3 = k + 2 \Rightarrow 2 k + 4 = 3 k - 3 a n d 9 = k + 2 \Rightarrow k = 7 a n d k = 7$

Therefore, the given system of equations will have infinitely many solutions, if k = 7.

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