Question

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

$\Huge\mathfrak\purple{answer:}$

$\bf \frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3x - 1} \\ \\ \bf \implies \frac{(3 \times 2) -(x + 1)}{2(x + 1)} = \frac{2}{3x - 1} \\ \\ \bf \implies \frac{6 - x - 1}{2x + 2} = \frac{2}{3x - 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf \implies \frac{5 - x}{2x + 2} = \frac{2}{3x - 1} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf \implies (5 - x)(3x - 1) = 2(2x + 2) \\ \\ \bf \implies 15x - 5 - {3x}^{2} + x = 4x + 4 \\ \\ \bf \implies 16x - 4x - {3x}^{2} = 4 + 5 \: \: \: \: \: \: \: \\ \\ \bf \implies 12x - {3x}^{2} = 9 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf \implies - {3x}^{2} + 12x - 9 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\huge\sf\purple{Hope~it~helps..!!}$

Answer 2

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

Given:

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

To find:

• Value of x.

Solution:

$\implies \dfrac{3}{x + 1} - \dfrac{1}{2} = \dfrac{2}{3x - 1} \\ \\ \implies \dfrac{3*2 - 1(x + 1)}{2(x + 1)} = \dfrac{2}{3x - 1} \\ \\ \implies \dfrac{6 - x - 1}{2x + 2} = \dfrac{2}{3x - 1} \\ \\ \implies (5 - x)(3x - 1) = 2*(2x + 2) \\ \\ \implies 16x - 3x^2 - 5 = 4x + 4 \\ \\ \implies 3x^2 - 16x + 5 + 4x + 4 = 0 \\ \\ \implies 3x^2 - 12x + 9 = 0 \\ \\ \implies x^2 - 4x + 3 = 0 \\ \\ \implies x^2 - x - 3x + 3 = 0 \\ \\ \implies x(x - 1) - 3(x - 1) = 0 \\ \\ \implies (x - 1)*(x - 3) = 0 \\ \\ \implies \text{x - 1 = 0 OR x - 3 = 0 } \\ \\ \implies \textbf{x = 1 OR x = 3 }$

HOPE IT HELPS!!!!