Question

Using factorization l, solve the quadratic equation:

$2( \frac{2x - 1}{x + 3}) - 3( \frac{x + 3}{2x - 1} ) = 5$

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Answer 1

$\large{\mathfrak{Answer -}}$

To prove,

$2( \frac{2x - 1}{x + 3} ) - 3( \frac{x + 3}{2x - 1} ) = 5$
Refer to the attachment !!..

Answer 2

$<b><font color = "red">Hello Friend$


Find your answer below ⬇⬇


$\mathsf{\implies 2 ({\dfrac {2x - 1}{x + 3}}) - 3({\dfrac{x + 3}{2x - 1}}) = 5}$


$\mathsf {\implies {\dfrac {2 (2x - 1)^{2} - 3 (x + 3)^{2}}{(x + 3)(2x - 1) }} = 5}$


$\mathsf {\implies {\dfrac {2 ( 4x^{2} + 1 - 4x) - 3(x^{2} + 9 + 6x)}{2x^{2} - x + 6x - 3}} = 5}$


$\mathsf{\implies {\dfrac{8x^{2} + 2 - 8x - 3x^{2} - 27 - 18x}{2x^{2} + 5x - 3}} = 5}$


$\mathsf{\implies 5x^{2} - 25 - 26x = 10x^{2} + 25x - 15}$


$\mathsf {\implies 5x^{2} - 25 - 26x - 10x^{2} - 25x + 15 = 0}$


$\mathsf{\implies - 5x^{2} - 51x - 10 = 0}$


$\mathsf{\implies -5x^{2} - 50x - x - 10 = 0}$


$\mathsf{\implies -5x(x + 10) - 1(x + 10) = 0}$


$\mathsf {\implies (- 5x - 1)(x + 10) = 0}$


$\mathsf {\implies -5x - 1 = 0}$


$\mathsf{\implies x = {\dfrac {- 1}{5}}}$


$\mathsf{\implies x + 10 = 0}$


$\mathsf{\implies x = - 10}$


$\mathsf{\implies x = {\dfrac {- 1}{5}} \: and - 10}$


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