Question

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

Answer 1

Answer:

$\huge{\bold☘}\mathfrak\pink{\bold{\underline{{ ℘ɧεŋσɱεŋศɭ}}}}{\bold☘}$

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

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

$\huge\tt{\boxed{\overbrace{\underbrace{\blue{Answer</p><p> }}}}}$

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_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

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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_____________________

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Answer 2

Answer:

\huge{\bold☘}\mathfrak\pink{\bold{\underline{{ ℘ɧεŋσɱεŋศɭ}}}}{\bold☘}☘

℘ɧεŋσɱεŋศɭ

☘

\huge\tt\red{\bold{\underline{\underline{❥Question᎓}}}}

❥Question᎓

solve the equation for x :

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

2x+1

4x−3

)−10(

4x−3

2x+1

)=3

\huge\tt{\boxed{\overbrace{\underbrace{\blue{Answer }}}}}

Answer

╔════════════════════════╗

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

GIVEN EQUATION IS

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

❲

2x+1

4x−3

❳−10(

4x−3

2x+1

)=3

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

2x+1

4x−3

)−10(

4x−3

2x+1

)=3

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

(2x+1)(4x−3)

(4x−3)

2

−10(2x+1)

2

=3

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

2

−24x+9)−10(4x

2

+4x+1)

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

2

−6x+4x−3)

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

2

−24x+9−40x

2

−40x−10

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

2

−18x+12x−9

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

2

−64x−1=24x

2

−6x−9

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

2

−24x

2

−64x+6x−1+9=0

⟹\bold{- 48 {x}^{2} - 58x + 8 = 0}⟹−48x

2

−58x+8=0

⟹\bold{24 {x}^{2} + 29x - 4 = 0}⟹24x

2

+29x−4=0

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

2

+32x−3x−4=0

⟹\bold{8x(3x + 4) - 1(3x + 4) = 0}⟹8x(3x+4)−1(3x+4)=0

⟹\bold{(3x + 4)(8x - 1) = 0}⟹(3x+4)(8x−1)=0

⟹\bold{3x + 4 = 0}⟹3x+4=0

⟹\bold{8x - 1 = 0}⟹8x−1=0

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

x=−

3

4

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

x=

8

1

\bold{∴The\: solutions \:are\: -4/3\: and \:1/8}∴Thesolutionsare−4/3and1/8

╚════════════════════════╝

нσρє ıт нєłρs yσυ

_____________________

тнαηkyσυ