Question

$\tt \:  \longrightarrow \: \dfrac{x + 3}{x - 2} - \dfrac{1 - x}{x} = \dfrac{17}{4}$

Find x​

Answer 1

To Find :-

• Value of "x"

Solution :-

$\sf:\implies \: \dfrac{x + 3}{x - 2} - \dfrac{1 - x}{x} = \dfrac{17}{4}$

$\sf:\implies \: \dfrac{(x + 3)x - (x - 2)(1 - x)}{(x - 2)x} = \dfrac{17}{4}$

$\sf:\implies \: \dfrac{ {x}^{2} + 3x - (x - {x}^{2} - 2 + 2x) }{ {x}^{2} - 2x} = \dfrac{17}{4}$

$\sf:\implies \: \dfrac{ {x}^{2} + 3x - 3x + 2 + {x}^{2} }{ {x}^{2} - 2x} = \dfrac{17}{4}$

$\sf:\implies  \: \dfrac{2 {x}^{2} + 2}{ {x}^{2} - 2x} = \dfrac{17}{4}$

$\sf:\implies \: {8x}^{2} + 8 = {17x}^{2} - 34x$

$\sf:\implies \: {9x}^{2} - 34x - 8 = 0$

$\sf:\implies \: {9x}^{2} - 36x + 2x - 8 = 0$

$\sf:\implies  \: 9x(x - 4) + 2(x - 4) = 0$

$\sf:\implies \: (x - 4)(9x + 2) = 0$

$\sf\implies \: \boxed{ \red {\sf\:  x \: = 4 \: OR\: x \: = \: - \: \dfrac{2}{9} }}$

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