What is the answer to (x^{-1}+A)^{-1}=B ?

The answer to (x^{-1}+A)^{-1}=B is x= B/(1-AB);\quad A\ne 1/B

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(x^{-1}+A)^{-1}=B

(x^{- 1} + A)^{- 1} = B

x = \frac{B}{1 - A B}; A \neq = \frac{1}{B}

Solution steps

(x^{- 1} + A)^{- 1} = B

Refine

\frac{x}{1 + A x} = B

Multiply both sides by

1 + A x

x = B (1 + A x)

Expand

B (1 + A x) : B + A B x

x = B + A B x

Move

A B x

to the left side

x - A B x = B

Factor

x - A B x : x (1 - A B)

x (1 - A B) = B

Divide both sides by

1 - A B; A \neq = \frac{1}{B}

x = \frac{B}{1 - A B}; A \neq = \frac{1}{B}

\frac{2}{3 x} - \frac{7}{5 x} = 1

a = \frac{1}{( 1 + x )}

\frac{- 3}{x - 3} + \frac{1}{x + 5} = 1

\frac{2 π}{b} = 6

\frac{1}{2 ( h - 6 )} = 0

What is the answer to (x^{-1}+A)^{-1}=B ?
The answer to (x^{-1}+A)^{-1}=B is x= B/(1-AB);\quad A\ne 1/B