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Solve the equation: $\frac{8}{2 x - 1} + \frac{9}{3 x - 1} = \frac{7}{x + 1}$ (Equation)

Created: 1 year ago | Updated: 1 year ago

Updated: 1 year ago

$\frac{8}{2 x - 1} + \frac{9}{3 x - 1} = \frac{7}{x + 1} \Rightarrow \frac{8}{2 x - 1} + \frac{9}{3 x - 1} = \frac{4}{x + 1} + \frac{3}{x + 1} \Rightarrow \frac{8}{2 x - 1} - \frac{4}{x + 1} = \frac{3}{x + 1} - \frac{9}{3 x - 1} \Rightarrow \frac{8 (x + 1) - 4 (2 x - 1)}{(2 x - 1) (x + 1)} = \frac{3 (3 x - 1) - 9 (x + 1)}{(x + 1) (3 x - 1)} \Rightarrow \frac{8 x + 8 - 8 x + 4}{(2 x - 1) (x + 1)} = \frac{9 x - 3 - 9 x - 9}{(x + 1) (3 x - 1)}$

$\Rightarrow \frac{12}{2 x - 1} = \frac{- 12}{3 x - 1}$ (উভয় পক্ষকে (x + 1) দ্বারা গুণ করে ]

$\Rightarrow \frac{1}{2 x - 1} = \frac{- 1}{3 x - 1}$ [উভয় পক্ষকে 12 দ্বারা ভাগ করে ]

$\Rightarrow 3 x - 1 = - 2 x + 1 \Rightarrow 3 x + 2 x = 1 + 1 \Rightarrow 5 x = 2 ∴ x = \frac{2}{5}$

1 year ago

NRBC Bank Ltd. || Management Trainee Officer (19-10-2018) গণিত

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