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

2x+3y-7=0
(k−1)x+(k+2)y-3k=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, $a_{1}$ =2, $b_{1}$ =3, $c_{1}$ =−7

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

For unique solution,we have

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
$\frac{2}{k - 1} = \frac{3}{k + 2} = \frac{- 7}{- 3 k}$

$\frac{2}{k - 1} = \frac{3}{k + 2}$ and $\frac{3}{k + 2} = \frac{- 7}{- 3 k}$

⇒2k+4=3k−3 and 9k=7k+14
⇒k=7and k=7

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

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