Find the Inverse (-2x-1)/(x+5)

$\frac{- 2 x - 1}{x + 5}$

Step 1

Interchange the variables.

$x = \frac{- 2 y - 1}{y + 5}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{- 2 y - 1}{y + 5} = x$ .

$\frac{- 2 y - 1}{y + 5} = x$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y + 5, 1$

Remove parentheses.

$y + 5, 1$

The LCM of one and any expression is the expression.

$y + 5$

Multiply each term in $\frac{- 2 y - 1}{y + 5} = x$ by $y + 5$ to eliminate the fractions.

Multiply each term in $\frac{- 2 y - 1}{y + 5} = x$ by $y + 5$ .

$\frac{- 2 y - 1}{y + 5} (y + 5) = x (y + 5)$

Simplify the left side.

Cancel the common factor of $y + 5$ .

Cancel the common factor.

$\frac{- 2 y - 1}{y + 5} (y + 5) = x (y + 5)$

Rewrite the expression.

$- 2 y - 1 = x (y + 5)$

Simplify the right side.

Apply the distributive property.

$- 2 y - 1 = x y + x \cdot 5$

Move $5$ to the left of $x$ .

$- 2 y - 1 = x y + 5 x$

Solve the equation.

Subtract $x y$ from both sides of the equation.

$- 2 y - 1 - x y = 5 x$

Add $1$ to both sides of the equation.

$- 2 y - x y = 5 x + 1$

Factor $y$ out of $- 2 y - x y$ .

Factor $y$ out of $- 2 y$ .

$y \cdot - 2 - x y = 5 x + 1$

Factor $y$ out of $- x y$ .

$y \cdot - 2 + y (- x) = 5 x + 1$

Factor $y$ out of $y \cdot - 2 + y (- x)$ .

$y (- 2 - x) = 5 x + 1$

Divide each term in $y (- 2 - x) = 5 x + 1$ by $- 2 - x$ and simplify.

Divide each term in $y (- 2 - x) = 5 x + 1$ by $- 2 - x$ .

$\frac{y (- 2 - x)}{- 2 - x} = \frac{5 x}{- 2 - x} + \frac{1}{- 2 - x}$

Simplify the left side.

Cancel the common factor of $- 2 - x$ .

Cancel the common factor.

$\frac{y (- 2 - x)}{- 2 - x} = \frac{5 x}{- 2 - x} + \frac{1}{- 2 - x}$

Divide $y$ by $1$ .

$y = \frac{5 x}{- 2 - x} + \frac{1}{- 2 - x}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{5 x + 1}{- 2 - x}$

Rewrite $- 2$ as $- 1 (2)$ .

$y = \frac{5 x + 1}{- 1 (2) - x}$

Factor $- 1$ out of $- x$ .

$y = \frac{5 x + 1}{- 1 (2) - (x)}$

Factor $- 1$ out of $- 1 (2) - (x)$ .

$y = \frac{5 x + 1}{- 1 (2 + x)}$

Move the negative in front of the fraction.

$y = - \frac{5 x + 1}{2 + x}$

Step 3

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

$f^{- 1} (x) = - \frac{5 x + 1}{2 + x}$

Step 4

Verify if $f^{- 1} (x) = - \frac{5 x + 1}{2 + x}$ is the inverse of $f (x) = \frac{- 2 x - 1}{x + 5}$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (\frac{- 2 x - 1}{x + 5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{5 (\frac{- 2 x - 1}{x + 5}) + 1}{2 + \frac{- 2 x - 1}{x + 5}}$

Remove parentheses.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{5 (\frac{- 2 x - 1}{x + 5}) + 1}{2 + \frac{- 2 x - 1}{x + 5}}$

Multiply the numerator and denominator of the complex fraction by $x + 5$ .

Multiply $\frac{5 \frac{- 2 x - 1}{x + 5} + 1}{2 + \frac{- 2 x - 1}{x + 5}}$ by $\frac{x + 5}{x + 5}$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - (\frac{x + 5}{x + 5} \cdot \frac{5 (\frac{- 2 x - 1}{x + 5}) + 1}{2 + \frac{- 2 x - 1}{x + 5}})$

Combine.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (5 (\frac{- 2 x - 1}{x + 5}) + 1)}{(x + 5) (2 + \frac{- 2 x - 1}{x + 5})}$

Apply the distributive property.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (5 (\frac{- 2 x - 1}{x + 5})) + (x + 5) \cdot 1}{(x + 5) \cdot 2 + (x + 5) (\frac{- 2 x - 1}{x + 5})}$

Cancel the common factor of $x + 5$ .

Cancel the common factor.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (5 (\frac{- 2 x - 1}{x + 5})) + (x + 5) \cdot 1}{(x + 5) \cdot 2 + (x + 5) (\frac{- 2 x - 1}{x + 5})}$

Rewrite the expression.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (5 (\frac{- 2 x - 1}{x + 5})) + (x + 5) \cdot 1}{(x + 5) \cdot 2 - 2 x - 1}$

Simplify the numerator.

Factor $x + 5$ out of $(x + 5) \cdot 5 \frac{- 2 x - 1}{x + 5} + (x + 5) \cdot 1$ .

Factor $x + 5$ out of $(x + 5) \cdot 5 \frac{- 2 x - 1}{x + 5}$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (5 (\frac{- 2 x - 1}{x + 5})) + (x + 5) \cdot 1}{(x + 5) \cdot 2 - 2 x - 1}$

Factor $x + 5$ out of $(x + 5) \cdot 1$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (5 (\frac{- 2 x - 1}{x + 5})) + (x + 5) (1)}{(x + 5) \cdot 2 - 2 x - 1}$

Factor $x + 5$ out of $(x + 5) (5 \frac{- 2 x - 1}{x + 5}) + (x + 5) (1)$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (5 (\frac{- 2 x - 1}{x + 5}) + 1)}{(x + 5) \cdot 2 - 2 x - 1}$

Combine $5$ and $\frac{- 2 x - 1}{x + 5}$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (\frac{5 (- 2 x - 1)}{x + 5} + 1)}{(x + 5) \cdot 2 - 2 x - 1}$

Write $1$ as a fraction with a common denominator.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (\frac{5 (- 2 x - 1)}{x + 5} + \frac{x + 5}{x + 5})}{(x + 5) \cdot 2 - 2 x - 1}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (\frac{5 (- 2 x - 1) + x + 5}{x + 5})}{(x + 5) \cdot 2 - 2 x - 1}$

Reorder terms.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (\frac{x + 5 (- 2 x - 1) + 5}{x + 5})}{(x + 5) \cdot 2 - 2 x - 1}$

Rewrite $\frac{x + 5 (- 2 x - 1) + 5}{x + 5}$ in a factored form.

Apply the distributive property.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (\frac{x + 5 (- 2 x) + 5 \cdot - 1 + 5}{x + 5})}{(x + 5) \cdot 2 - 2 x - 1}$

Multiply $- 2$ by $5$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (\frac{x - 10 x + 5 \cdot - 1 + 5}{x + 5})}{(x + 5) \cdot 2 - 2 x - 1}$

Multiply $5$ by $- 1$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (\frac{x - 10 x - 5 + 5}{x + 5})}{(x + 5) \cdot 2 - 2 x - 1}$

Subtract $10 x$ from $x$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (\frac{- 9 x - 5 + 5}{x + 5})}{(x + 5) \cdot 2 - 2 x - 1}$

Add $- 5$ and $5$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (\frac{- 9 x + 0}{x + 5})}{(x + 5) \cdot 2 - 2 x - 1}$

Add $- 9 x$ and $0$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (\frac{- 9 x}{x + 5})}{(x + 5) \cdot 2 - 2 x - 1}$

Move the negative in front of the fraction.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) (- \frac{(9) x}{x + 5})}{(x + 5) \cdot 2 - 2 x - 1}$

Remove unnecessary parentheses.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{(x + 5) \cdot (- 1 \frac{9 x}{x + 5})}{(x + 5) \cdot 2 - 2 x - 1}$

Factor out negative.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{- (x + 5) \frac{9 x}{x + 5}}{(x + 5) \cdot 2 - 2 x - 1}$

Simplify the denominator.

Apply the distributive property.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{- (x + 5) \frac{9 x}{x + 5}}{x \cdot 2 + 5 \cdot 2 - 2 x - 1}$

Move $2$ to the left of $x$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{- (x + 5) \frac{9 x}{x + 5}}{2 \cdot x + 5 \cdot 2 - 2 x - 1}$

Multiply $5$ by $2$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{- (x + 5) \frac{9 x}{x + 5}}{2 \cdot x + 10 - 2 x - 1}$

Subtract $2 x$ from $2 x$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{- (x + 5) \frac{9 x}{x + 5}}{0 + 10 - 1}$

Add $0$ and $10$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{- (x + 5) \frac{9 x}{x + 5}}{10 - 1}$

Subtract $1$ from $10$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - \frac{- (x + 5) \frac{9 x}{x + 5}}{9}$

Reduce the expression by cancelling the common factors.

Factor $\frac{9 x}{x + 5}$ out of $\frac{- (x + 5) \frac{9 x}{x + 5}}{9}$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - (\frac{9 x}{x + 5} \cdot \frac{- (x + 5)}{9})$

Cancel the common factor of $9$ .

Factor $9$ out of $9 x$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - (\frac{9 (x)}{x + 5} \cdot \frac{- (x + 5)}{9})$

Cancel the common factor.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - (\frac{9 x}{x + 5} \cdot \frac{- (x + 5)}{9})$

Rewrite the expression.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - (\frac{x}{x + 5} \cdot (- (x + 5)))$

Cancel the common factor of $x + 5$ .

Factor $x + 5$ out of $- (x + 5)$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - (\frac{x}{x + 5} \cdot ((x + 5) \cdot - 1))$

Cancel the common factor.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - (\frac{x}{x + 5} \cdot ((x + 5) \cdot - 1))$

Rewrite the expression.

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - (x \cdot - 1)$

Simplify the expression.

Move $- 1$ to the left of $x$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = - (- 1 \cdot x)$

Rewrite $- 1 x$ as $- x$ .

$f^{- 1} (\frac{- 2 x - 1}{x + 5}) = x$

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (- \frac{5 x + 1}{2 + x})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{- 2 (- \frac{5 x + 1}{2 + x}) - 1}{(- \frac{5 x + 1}{2 + x}) + 5}$

Multiply the numerator and denominator of the complex fraction by $2 + x$ .

Multiply $\frac{- 2 (- \frac{5 x + 1}{2 + x}) - 1}{- \frac{5 x + 1}{2 + x} + 5}$ by $\frac{2 + x}{2 + x}$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{2 + x}{2 + x} \cdot \frac{- 2 (- \frac{5 x + 1}{2 + x}) - 1}{- \frac{5 x + 1}{2 + x} + 5}$

Combine.

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- 2 (- \frac{5 x + 1}{2 + x}) - 1)}{(2 + x) (- \frac{5 x + 1}{2 + x} + 5)}$

Apply the distributive property.

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- 2 (- \frac{5 x + 1}{2 + x})) + (2 + x) \cdot - 1}{(2 + x) (- \frac{5 x + 1}{2 + x}) + (2 + x) \cdot 5}$

Cancel the common factor of $2 + x$ .

Move the leading negative in $- \frac{5 x + 1}{2 + x}$ into the numerator.

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- 2 (- \frac{5 x + 1}{2 + x})) + (2 + x) \cdot - 1}{(2 + x) (\frac{- (5 x + 1)}{2 + x}) + (2 + x) \cdot 5}$

Cancel the common factor.

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- 2 (- \frac{5 x + 1}{2 + x})) + (2 + x) \cdot - 1}{(2 + x) (\frac{- (5 x + 1)}{2 + x}) + (2 + x) \cdot 5}$

Rewrite the expression.

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- 2 (- \frac{5 x + 1}{2 + x})) + (2 + x) \cdot - 1}{- (5 x + 1) + (2 + x) \cdot 5}$

Simplify the numerator.

Factor $2 + x$ out of $(2 + x) (- 2 (- \frac{5 x + 1}{2 + x})) + (2 + x) \cdot - 1$ .

Factor $2 + x$ out of $(2 + x) \cdot - 1$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- 2 (- \frac{5 x + 1}{2 + x})) + (2 + x) (- 1)}{- (5 x + 1) + (2 + x) \cdot 5}$

Factor $2 + x$ out of $(2 + x) (- 2 (- \frac{5 x + 1}{2 + x})) + (2 + x) (- 1)$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- 2 (- \frac{5 x + 1}{2 + x}) - 1)}{- (5 x + 1) + (2 + x) \cdot 5}$

Factor $- 1$ out of $- 2 (- \frac{5 x + 1}{2 + x}) - 1$ .

Reorder $2$ and $x$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- 2 \cdot (- 1 \frac{5 x + 1}{x + 2}) - 1)}{- (5 x + 1) + (2 + x) \cdot 5}$

Factor $- 1$ out of $- 2 \cdot - 1 \frac{5 x + 1}{x + 2}$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- (- 2 \frac{5 x + 1}{x + 2}) - 1)}{- (5 x + 1) + (2 + x) \cdot 5}$

Rewrite $- 1$ as $- 1 (1)$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- (- 2 \frac{5 x + 1}{x + 2}) - 1 \cdot 1)}{- (5 x + 1) + (2 + x) \cdot 5}$

Factor $- 1$ out of $- (- 2 \frac{5 x + 1}{x + 2}) - 1 (1)$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- (- 2 \frac{5 x + 1}{x + 2} + 1))}{- (5 x + 1) + (2 + x) \cdot 5}$

Combine $- 2$ and $\frac{5 x + 1}{x + 2}$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- (\frac{- 2 (5 x + 1)}{x + 2} + 1))}{- (5 x + 1) + (2 + x) \cdot 5}$

Move the negative in front of the fraction.

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- (- \frac{(2) (5 x + 1)}{x + 2} + 1))}{- (5 x + 1) + (2 + x) \cdot 5}$

Write $1$ as a fraction with a common denominator.

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- (- \frac{2 (5 x + 1)}{x + 2} + \frac{x + 2}{x + 2}))}{- (5 x + 1) + (2 + x) \cdot 5}$

Combine the numerators over the common denominator.

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- \frac{- 2 (5 x + 1) + x + 2}{x + 2})}{- (5 x + 1) + (2 + x) \cdot 5}$

Reorder terms.

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- \frac{x + 2 - 2 (5 x + 1)}{x + 2})}{- (5 x + 1) + (2 + x) \cdot 5}$

Rewrite $\frac{x + 2 - 2 (5 x + 1)}{x + 2}$ in a factored form.

Apply the distributive property.

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- \frac{x + 2 - 2 (5 x) - 2 \cdot 1}{x + 2})}{- (5 x + 1) + (2 + x) \cdot 5}$

Multiply $5$ by $- 2$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- \frac{x + 2 - 10 x - 2 \cdot 1}{x + 2})}{- (5 x + 1) + (2 + x) \cdot 5}$

Multiply $- 2$ by $1$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- \frac{x + 2 - 10 x - 2}{x + 2})}{- (5 x + 1) + (2 + x) \cdot 5}$

Subtract $10 x$ from $x$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- \frac{- 9 x + 2 - 2}{x + 2})}{- (5 x + 1) + (2 + x) \cdot 5}$

Subtract $2$ from $2$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- \frac{- 9 x + 0}{x + 2})}{- (5 x + 1) + (2 + x) \cdot 5}$

Add $- 9 x$ and $0$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (- \frac{- 9 x}{x + 2})}{- (5 x + 1) + (2 + x) \cdot 5}$

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) \cdot (- 1 \frac{- 9 x}{x + 2})}{- (5 x + 1) + (2 + x) \cdot 5}$

Factor out negative.

$f (- \frac{5 x + 1}{2 + x}) = \frac{- ((2 + x) (\frac{- 9 x}{x + 2}))}{- (5 x + 1) + (2 + x) \cdot 5}$

Move the negative in front of the fraction.

$f (- \frac{5 x + 1}{2 + x}) = \frac{- ((2 + x) (- \frac{(9) x}{x + 2}))}{- (5 x + 1) + (2 + x) \cdot 5}$

Remove unnecessary parentheses.

$f (- \frac{5 x + 1}{2 + x}) = \frac{- (2 + x) \cdot (- 1 \frac{9 x}{x + 2})}{- (5 x + 1) + (2 + x) \cdot 5}$

Combine exponents.

Factor out negative.

$f (- \frac{5 x + 1}{2 + x}) = \frac{- ((2 + x) \cdot (- 1 \frac{9 x}{x + 2}))}{- (5 x + 1) + (2 + x) \cdot 5}$

Multiply $- 1$ by $- 1$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{1 ((2 + x) (\frac{9 x}{x + 2}))}{- (5 x + 1) + (2 + x) \cdot 5}$

Multiply $2 + x$ by $1$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (\frac{9 x}{x + 2})}{- (5 x + 1) + (2 + x) \cdot 5}$

Simplify the denominator.

Apply the distributive property.

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (\frac{9 x}{x + 2})}{- (5 x) - 1 \cdot 1 + (2 + x) \cdot 5}$

Multiply $5$ by $- 1$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (\frac{9 x}{x + 2})}{- 5 x - 1 \cdot 1 + (2 + x) \cdot 5}$

Multiply $- 1$ by $1$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (\frac{9 x}{x + 2})}{- 5 x - 1 + (2 + x) \cdot 5}$

Apply the distributive property.

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (\frac{9 x}{x + 2})}{- 5 x - 1 + 2 \cdot 5 + x \cdot 5}$

Multiply $2$ by $5$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (\frac{9 x}{x + 2})}{- 5 x - 1 + 10 + x \cdot 5}$

Move $5$ to the left of $x$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (\frac{9 x}{x + 2})}{- 5 x - 1 + 10 + 5 \cdot x}$

Add $- 5 x$ and $5 x$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (\frac{9 x}{x + 2})}{0 - 1 + 10}$

Subtract $1$ from $0$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (\frac{9 x}{x + 2})}{- 1 + 10}$

Add $- 1$ and $10$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{(2 + x) (\frac{9 x}{x + 2})}{9}$

Simplify terms.

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

$f (- \frac{5 x + 1}{2 + x}) = \frac{9 x}{x + 2} \cdot \frac{2 + x}{9}$

Cancel the common factor of $9$ .

Factor $9$ out of $9 x$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{9 (x)}{x + 2} \cdot \frac{2 + x}{9}$

Cancel the common factor.

$f (- \frac{5 x + 1}{2 + x}) = \frac{9 x}{x + 2} \cdot \frac{2 + x}{9}$

Rewrite the expression.

$f (- \frac{5 x + 1}{2 + x}) = \frac{x}{x + 2} \cdot (2 + x)$

Multiply $\frac{x}{x + 2}$ by $2 + x$ .

$f (- \frac{5 x + 1}{2 + x}) = \frac{x (2 + x)}{x + 2}$

Cancel the common factor of $2 + x$ and $x + 2$ .

Reorder terms.

$f (- \frac{5 x + 1}{2 + x}) = \frac{x (x + 2)}{x + 2}$

Cancel the common factor.

$f (- \frac{5 x + 1}{2 + x}) = \frac{x (x + 2)}{x + 2}$

Divide $x$ by $1$ .

$f (- \frac{5 x + 1}{2 + x}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{5 x + 1}{2 + x}$ is the inverse of $f (x) = \frac{- 2 x - 1}{x + 5}$ .

$f^{- 1} (x) = - \frac{5 x + 1}{2 + x}$

Find the Inverse (-2x-1)/(x+5)

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