Find the Inverse 18/x-9

$\frac{18}{x} - 9$

Interchange the variables.

$x = \frac{18}{y} - 9$

Solve for $y$ .

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Rewrite the equation as $\frac{18}{y} - 9 = x$ .

$\frac{18}{y} - 9 = x$

Add $9$ to both sides of the equation.

$\frac{18}{y} = x + 9$

Solve for $y$ .

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Multiply each term by $y$ and simplify.

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Multiply each term in $\frac{18}{y} = x + 9$ by $y$ .

$\frac{18}{y} \cdot y = x \cdot y + 9 \cdot y$

Cancel the common factor of $y$ .

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Cancel the common factor.

$\frac{18}{y} \cdot y = x \cdot y + 9 \cdot y$

Rewrite the expression.

$18 = x \cdot y + 9 \cdot y$

Multiply $9$ by $y$ .

$18 = x y + 9 y$

Rewrite the equation as $x y + 9 y = 18$ .

$x y + 9 y = 18$

Factor $y$ out of $x y + 9 y$ .

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Factor $y$ out of $x y$ .

$y x + 9 y = 18$

Factor $y$ out of $9 y$ .

$y x + y \cdot 9 = 18$

Factor $y$ out of $y x + y \cdot 9$ .

$y (x + 9) = 18$

Divide each term by $x + 9$ and simplify.

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Divide each term in $y (x + 9) = 18$ by $x + 9$ .

$\frac{y (x + 9)}{x + 9} = \frac{18}{x + 9}$

Cancel the common factor of $x + 9$ .

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Cancel the common factor.

$\frac{y (x + 9)}{x + 9} = \frac{18}{x + 9}$

Divide $y$ by $1$ .

$y = \frac{18}{x + 9}$

Solve for $y$ and replace with $f^{- 1} (x)$ .

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Replace the $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = \frac{18}{x + 9}$

Set up the composite result function.

$g (f (x))$

Evaluate $g (f (x))$ by substituting in the value of $f$ into $g$ .

$g (\frac{18}{x} - 9) = \frac{18}{(\frac{18}{x} - 9) + 9}$

Cancel the common factor of $18$ and $(\frac{18}{x} - 9) + 9$ .

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Factor $9$ out of $18$ .

$g (\frac{18}{x} - 9) = \frac{9 \cdot 2}{\frac{18}{x} - 9 + 9}$

Cancel the common factors.

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Factor $9$ out of $\frac{18}{x}$ .

$g (\frac{18}{x} - 9) = \frac{9 \cdot 2}{9 (\frac{2}{x}) - 9 + 9}$

Factor $9$ out of $- 9$ .

$g (\frac{18}{x} - 9) = \frac{9 \cdot 2}{9 (\frac{2}{x}) + 9 \cdot - 1 + 9}$

Factor $9$ out of $9 \frac{2}{x} + 9 \cdot - 1$ .

$g (\frac{18}{x} - 9) = \frac{9 \cdot 2}{9 (\frac{2}{x} - 1) + 9}$

Factor $9$ out of $9$ .

$g (\frac{18}{x} - 9) = \frac{9 \cdot 2}{9 (\frac{2}{x} - 1) + 9 (1)}$

Factor $9$ out of $9 (\frac{2}{x} - 1) + 9 (1)$ .

$g (\frac{18}{x} - 9) = \frac{9 \cdot 2}{9 (\frac{2}{x} - 1 + 1)}$

Cancel the common factor.

$g (\frac{18}{x} - 9) = \frac{9 \cdot 2}{9 (\frac{2}{x} - 1 + 1)}$

Rewrite the expression.

$g (\frac{18}{x} - 9) = \frac{2}{\frac{2}{x} - 1 + 1}$

Simplify the denominator.

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Add $- 1$ and $1$ .

$g (\frac{18}{x} - 9) = \frac{2}{\frac{2}{x} + 0}$

Add $\frac{2}{x}$ and $0$ .

$g (\frac{18}{x} - 9) = \frac{2}{\frac{2}{x}}$

Multiply the numerator by the reciprocal of the denominator.

$g (\frac{18}{x} - 9) = 2 (\frac{x}{2})$

Cancel the common factor of $2$ .

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Cancel the common factor.

$g (\frac{18}{x} - 9) = 2 (\frac{x}{2})$

Rewrite the expression.

$g (\frac{18}{x} - 9) = x$

Since $g (f (x)) = x$ , $f^{- 1} (x) = \frac{18}{x + 9}$ is the inverse of $f (x) = \frac{18}{x} - 9$ .

$f^{- 1} (x) = \frac{18}{x + 9}$

Find the Inverse 18/x-9

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