Question

$\frac{3}{x + 1 } - \frac{1}{2} = \frac{2}{3x - 1}$
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Answer 1

For solution, please refer to the attachment.

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Value\:of\:x=1\:and\:3}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies \frac{3}{x + 1 } - \frac{1}{2} = \frac{2}{3x - 1} \\ \\ \red{\underline \bold{To \: Find :}}\\ \tt: \implies value \: of \: x = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\\tt: \implies \frac{3}{x + 1 } - \frac{1}{2} = \frac{2}{3x - 1} \\ \\ \tt: \implies \frac{3 \times 2 - (x + 1)}{2(x + 1)} = \frac{2}{3x - 1} \\ \\\tt: \implies \frac{6 - x - 1}{2x + 2} = \frac{2}{3x - 1} \\ \\ \tt: \implies \frac{5 - x}{2x + 2} = \frac{2}{3x - 1} \\ \\\tt: \implies (5 - x)(3x - 1) = 4x + 4 \\ \\\tt: \implies 15x - 3 {x}^{2} - 5 + x = 4x + 4 \\ \\\tt: \implies 16x - 4x - 3 {x}^{2} - 5 - 4 = 0 \\ \\ \tt: \implies 12x - {3x}^{2} - 9 = 0 \\ \\\tt: \implies 3 {x}^{2} - 12x + 9 = 0 \\ \\\tt: \implies {x}^{2} - 4x + 3 = 0 \\ \\\tt: \implies {x}^{2} - 3x - x + 3 = 0 \\ \\\tt: \implies x(x - 3) - 1(x - 3) = 0 \\ \\\tt: \implies (x - 1)(x - 3) = 0 \\ \\ \green{\tt: \implies x = 1 \: and \: 3}$