Question

$3 \div( x + 1 ) - 1 \div 2 = 2 \div (3x + 1)$
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Answer 1

$\huge \tt \frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3x + 1} \\ \\ \\ \bf = > \frac{6 - (x + 1)}{2(x + 1)} = \frac{2}{3x + 1} \\ \\ \bf = > \frac{6 - x - 1}{2x + 2} = \frac{2}{3x + 1} \\ \\ \bf = > \frac{5 - x}{2x + 2} = \frac{2}{3x + 1} \\ \\ \bf = > 15x + 5 - {3x}^{2} - x = 4x + 4 \\ \bf = > { - 3x}^{2} + 5 + 14x - 4x - 4 = 0 \\ \bf = > { - 3x}^{2} + 10x + 1 = 0$

Now,

$\bf{D = {b}^{2} - 4ac} \\ = > {(10)}^{2} - 2( - 3)(1) \\ = > 100 + 6 \: \: \: \: = 106$

Then,

$\bf\large{x = \frac{ - b \pm \sqrt{D} }{2a} }\\ \\ \sf{ x = > \frac{ - 10 \pm \sqrt{106} }{2( - 3)}} \\ \\ \sf{x = > \frac{ - 10 + \sqrt{106} }{ - 6} \: \: \: \: \: or \: \: \: \frac{ - 10 - \sqrt{106} }{ - 6} } \\ \\ \boxed{ \boxed{ x = > \frak{ \red{ \frac{10 - \sqrt{106} }{6}} \: \: \: \: \: \: { \sf{or}} \: \: \: \: \red{ \frac{10 + \sqrt{106} }{6}} }}}$