What is the answer to AX(D+BX)^{-1}=C ?

The answer to AX(D+BX)^{-1}=C is X=(DC)/(A-BC);\quad A\ne BC

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AX(D+BX)^{-1}=C

A X (D + BX)^{- 1} = C

X = \frac{D C}{A - BC}; A \neq = BC

Solution steps

A X (D + BX)^{- 1} = C

Refine

\frac{A X}{D + BX} = C

Multiply both sides by

D + BX

A X = C (D + BX)

Expand

C (D + BX) : D C + BCX

A X = D C + BCX

Move

BCX

to the left side

A X - BCX = D C

Factor

A X - BCX : X (A - BC)

X (A - BC) = D C

Divide both sides by

A - BC; A \neq = BC

X = \frac{D C}{A - BC}; A \neq = BC

f, f (x)^{- 1} (x) = x^{3} + x

\frac{2 x - 3}{x - 1} = \frac{8 x + 1}{4 x + 5}

\frac{1}{5 x} + \frac{x + 4}{x} = \frac{6}{5}

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

\frac{4}{x ^{2}} = 0

What is the answer to AX(D+BX)^{-1}=C ?
The answer to AX(D+BX)^{-1}=C is X=(DC)/(A-BC);\quad A\ne BC