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Solve for x : `1/(2a + b + 2x) =1/(2a) + 1/b + 1/(2x); x ≠ 0, x ≠ (−2a −b)/2`, a, b ≠ 0

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#### Solution

`1/(2a+b+2x) =1/(2a) + 1/b + 1/(2x)`

`1/(2a+b+2x) −1/(2x) =1/(2a) +1/b`

`(2x−2a−b−2x)/(4ax+2bx+4x^2) =(b+2a)/(2ab)`

`(−2a−b)(2ab)=(b+2a)(4ax+2bx+4x^2)`

`(−(b+2a)(2ab))/(b+2a)=(4ax+2bx+4x^2)`

− (2ab) = 4ax + 2bx + 4x^{2}

4x^{2} + 2bx + 4ax + 2ab = 0

2x( 2x + b ) + 2a( 2x + b) = 0

(2x + 2a)(2x + b) = 0

⇒ (2x + 2a) = 0

⇒ x = − a

or

(2x + b) = 0

⇒ x = `− b/2`

Therefore, values of x are − a and `− b/2`.

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