What is the answer to solvefor a,(a+b)^{-1}=a^{-1}+b^{-1} ?

The answer to solvefor a,(a+b)^{-1}=a^{-1}+b^{-1} is a=-b/2+i(sqrt(3)b)/2 ,a=-b/2-i(sqrt(3)b)/2

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solvefor a,(a+b)^{-1}=a^{-1}+b^{-1}

solve for

a, (a + b)^{- 1} = a^{- 1} + b^{- 1}

a = - \frac{b}{2} + i \frac{3 b}{2}, a = - \frac{b}{2} - i \frac{3 b}{2}

Solution steps

(a + b)^{- 1} = a^{- 1} + b^{- 1}

Refine

\frac{1}{a + b} = \frac{1}{a} + \frac{1}{b}

Multiply by LCM

ab = b (a + b) + a (a + b)

Solve

ab = b (a + b) + a (a + b) : a = - \frac{b}{2} + i \frac{3 b}{2}, a = - \frac{b}{2} - i \frac{3 b}{2}

a = - \frac{b}{2} + i \frac{3 b}{2}, a = - \frac{b}{2} - i \frac{3 b}{2}

\frac{4}{x ^{2}} = \frac{1}{( a - x ) 2}

\frac{12}{23} = \frac{0.5}{x}

u, \frac{1}{u} + \frac{1}{v} = \frac{1}{f}

2 x = \frac{1}{3 x ^{2}}

\frac{x + 1}{x} = \frac{29}{10}

What is the answer to solvefor a,(a+b)^{-1}=a^{-1}+b^{-1} ?
The answer to solvefor a,(a+b)^{-1}=a^{-1}+b^{-1} is a=-b/2+i(sqrt(3)b)/2 ,a=-b/2-i(sqrt(3)b)/2