Question

solve the equation for x :
$\bold{ (\frac{4x - 3}{2x + 1} ) - 10 (\frac{2x + 1}{4x - 3} ) = 3}$


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Answer 1

$\huge\tt\red{\bold{\underline{\underline{❥Question᎓}}}}$solve the equation for x :

$\bold{ (\frac{4x - 3}{2x + 1} ) - 10 (\frac{2x + 1}{4x - 3} ) = 3}$

$\huge\mathcal{Answer}$

GIVEN EQUATION IS

$\bold{\boxed{\boxed{ ❲\frac{4x - 3}{2x + 1}❳ - 10 (\frac{2x + 1}{4x - 3} ) = 3}}}$

$⟹\bold{( \frac{4x - 3}{2x + 1} ) - 10 (\frac{2x + 1}{4x - 3} ) = 3}$

$⟹\bold{\frac{ {(4x - 3)}^{2} - 10 {(2x + 1)}^{2} }{(2x + 1)(4x - 3)} = 3}$

$\bold{⟹(16 {x}^{2} - 24x + 9) - 10(4 {x}^{2} + 4x + 1)}$

$\bold{= 3(8 {x}^{2} - 6x + 4x - 3)}$

$\bold{16 {x}^{2} - 24x + 9 - 40 {x}^{2} - 40x - 10}$

$\bold{ = 24 {x}^{2} - 18x + 12x - 9}$

$\bold{⟹- 24 {x}^{2} - 64x - 1 = 24 {x}^{2} - 6x - 9}$

$⟹\bold{- 24 {x}^{2} - 24 {x}^{2} - 64x + 6x - 1 + 9 = 0}$

$⟹\bold{- 48 {x}^{2} - 58x + 8 = 0}$

$⟹\bold{24 {x}^{2} + 29x - 4 = 0}$

$⟹\bold{24 {x}^{2} + 32x - 3x - 4 = 0}$

$⟹\bold{8x(3x + 4) - 1(3x + 4) = 0}$

$⟹\bold{(3x + 4)(8x - 1) = 0}$

$⟹\bold{3x + 4 = 0}$

$⟹\bold{8x - 1 = 0}$

$\bold{\boxed{\red{x = - \frac{4}{3} }}}$

$\bold{\boxed{\blue{x = \frac{1}{8}}}}$

$\bold{∴The\: solutions \:are\: -4/3\: and \:1/8}$

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