What value of K will be following pair of linear equations have infinite solutions
2x-3y=7
(K+1)x+(1-2K)y=(5K-4)

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Solution

Given , 2x-3y = 7 (1)
(k+1) x + (1-2k) y = 5k - 4 (2)
Here $a_{1} = 2, b_{1} = - 3, c_{1} = 7$
$a_{2} = k + 1, b_{2} = 1 - 2 k, c_{3} = 5 k - 4$
Since equations (1) and (2) have infinite solutions
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{3}}$
$\Rightarrow \frac{2}{k + 1} = \frac{- 3}{1 - 2 k} = \frac{7}{5 k - 4}$
$∴ \frac{2}{k + 1} = \frac{3}{1 - 2 k}$
$\Rightarrow 2 (1 - 2 k) = - 3 (k + 1)$
$\Rightarrow 2 - 4 k = - 3 k - 3$
$\Rightarrow 2 + 3 = - 3 k + 4 k$
$\Rightarrow k = 5$

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