Question

How to solve 4/(B+1)+4/(B-1)=3

Answer 1

$\textbf{ Hello Mate }$

4 / ( B + 1 ) + 4 / ( B - 1 ) = 3

[ 4( B - 1 ) + 4( B + 1 ) ] /(B+1)(B-1)

( 4B - 4 + 4B + 4 ) / ( B^2 - 1 ) = 3

8B = 3B^2 - 3

0 = 3B^2 - 8B - 3

0 = 3B^2 - 9B + B - 3

0 = 3B( B - 3 ) + 1 ( B - 3 )

0 = ( 3B + 1 )( B - 3 )

Each getting "0"

3B + 1 = 0 or B = - 1 / 3

B - 3 = 0 or B = 3

Hope it helps

Answer 2

$\dfrac{4}{B + 1} + \dfrac{4}{B - 1} = 3$

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Combine them into a single fraction:

$\dfrac{4(B - 1) + 4(B + 1) }{(B + 1)(B - 1)} = 3$

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Distribute the 4s / Apply (a + b) (a - b) = a² - b²:

$\dfrac{4B - 4 + 4B + 4 }{B^2 - 1} = 3$

.

Combine like terms / Cross multiply:

$8B = 3(B^2 - 1)$

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Distribute 3:

$8B = 3B^2 - 3$

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Take away 8B from both sides:

$3B^2- 8B - 3 = 0$

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Factorise:

$(b - 3)(3b + 1) = 0$

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Apply zero product property:

$b = 3 \ \ or \ \ b =-\dfrac{1}{3}$

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Answer: b = 3 or -1/3