Math, asked by Anonymous, 6 months ago

solve the equation for x:-
2 (\frac{2x - 1}{x + 3} ) - 3 ( \frac{x + 3}{2x - 1} ) = 5

Answers

Answered by Anonymous
99

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

\huge\tt\red{\bold{\underline{\underline{❥Question᎓}}}}\huge\tt\bold{2 (\frac{2x - 1}{x + 3} ) - 3 ( \frac{x + 3}{2x - 1} ) = 5}

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

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_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

GIVEN:-

\bold{\boxed{\boxed{\green{2 (\frac{2x - 1}{x + 3} ) - 3 ( \frac{x + 3}{2x - 1} ) = 5}}}}

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

 ⟹\bold{\frac{2 {(2x - 1) }^{2}  - 3 {(x + 3)}^{2} }{(x + 3)(2x - 1)}  = 5}

⟹\bold{2(4 {x}^{2}  - 4x + 1)  - 3( {x}^{2}  + 6x + 9)}

 \bold{= 5(2 {x}^{2}  + 6x - x - 3)}

⟹\bold{8 {x}^{2}  - 8x + 2 - 3 {x}^{2}  - 18x - 27}

\bold{ = 10 {x}^{2}  + 30x - 5x - 15}

⟹\bold{5 {x}^{2}  + 51x + 10 = 0}

⟹\bold{5 {x}^{2}  + 50x + x + 10 = 0}

⟹\bold{5x(x + 10) + 1(x + 10) = 0}

⟹\bold{(x + 10)(5x + 1) = 0}

⟹\bold{x + 10 = 0}

⟹\bold{5x + 1 = 0}

\bold{\boxed{\red{x =  - 10}}}

\bold{\boxed{x =   - \frac{1}{5}}}

\bold{\boxed{∴The \:solutions\: are \:-20\: and\: -1/5}}

╚════════════════════════╝

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_____________________

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Answered by Anonymous
0

Answer:

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

℘ɧεŋσɱεŋศɭ

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

❥Question᎓

\huge\tt\bold{2 (\frac{2x - 1}{x + 3} ) - 3 ( \frac{x + 3}{2x - 1} ) = 5}2(

x+3

2x−1

)−3(

2x−1

x+3

)=5

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

Answer

╔════════════════════════╗

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

GIVEN:-

\bold{\boxed{\boxed{\green{2 (\frac{2x - 1}{x + 3} ) - 3 ( \frac{x + 3}{2x - 1} ) = 5}}}}

2(

x+3

2x−1

)−3(

2x−1

x+3

)=5

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

x+3

2x−1

)−3(

2x−1

x+3

)=5

⟹\bold{\frac{2 {(2x - 1) }^{2} - 3 {(x + 3)}^{2} }{(x + 3)(2x - 1)} = 5}⟹

(x+3)(2x−1)

2(2x−1)

2

−3(x+3)

2

=5

⟹\bold{2(4 {x}^{2} - 4x + 1) - 3( {x}^{2} + 6x + 9)}⟹2(4x

2

−4x+1)−3(x

2

+6x+9)

\bold{= 5(2 {x}^{2} + 6x - x - 3)}=5(2x

2

+6x−x−3)

⟹\bold{8 {x}^{2} - 8x + 2 - 3 {x}^{2} - 18x - 27}⟹8x

2

−8x+2−3x

2

−18x−27

\bold{ = 10 {x}^{2} + 30x - 5x - 15}=10x

2

+30x−5x−15

⟹\bold{5 {x}^{2} + 51x + 10 = 0}⟹5x

2

+51x+10=0

⟹\bold{5 {x}^{2} + 50x + x + 10 = 0}⟹5x

2

+50x+x+10=0

⟹\bold{5x(x + 10) + 1(x + 10) = 0}⟹5x(x+10)+1(x+10)=0

⟹\bold{(x + 10)(5x + 1) = 0}⟹(x+10)(5x+1)=0

⟹\bold{x + 10 = 0}⟹x+10=0

⟹\bold{5x + 1 = 0}⟹5x+1=0

\bold{\boxed{\red{x = - 10}}}

x=−10

\bold{\boxed{x = - \frac{1}{5}}}

x=−

5

1

\bold{\boxed{∴The \:solutions\: are \:-20\: and\: -1/5}}

∴Thesolutionsare−20and−1/5

╚════════════════════════╝

нσρє ıт нєłρs yσυ

_____________________

тнαηkyσυ

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