Question

somebody please solve this question. wrong answers will be reported​

Answer 1

Step-by-step explanation:

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Answer 2

$Given \:(2a+3b)(2c-3d) = (2a-3b)(2c+3d)$

$\implies \frac{(2a+3b)}{2a-3b)} = \frac{(2c+3d)}{(2c-3d)}$

$\implies \frac{(2a+3b)\blue{+(2a-3b)}}{\blue{(2a-3b)}} = \frac{(2c+3d)\pink{ + (2c-3d)}}{\pink{(2c-3d)}}$

$\implies \frac{2a+3b+ 2a-3b}{2a-3b)} = \frac{2c+3d + 2c-3d}{(2c-3d)}$

$\implies \frac{4a}{2a-3b} = \frac{4c}{2c-3d}$

$\implies \frac{a}{2a-3b} = \frac{c}{2c-3d}$

$\implies a(2c-3d) = c(2a-3b)$

$\implies \cancel {2ac} - 3ad = \cancel {2ac} - 3bc$

$\implies -3ad = -3bc$

$\implies ad = bc$

$\implies \frac{a}{b} = \frac{c}{d}$

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