Find the Inverse (8x+9)/(x+8)

$\frac{8 x + 9}{x + 8}$

Step 1

Interchange the variables.

$x = \frac{8 y + 9}{y + 8}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{8 y + 9}{y + 8} = x$ .

$\frac{8 y + 9}{y + 8} = x$

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.

$y + 8, 1$

Remove parentheses.

$y + 8, 1$

The LCM of one and any expression is the expression.

$y + 8$

Multiply each term in $\frac{8 y + 9}{y + 8} = x$ by $y + 8$ to eliminate the fractions.

Multiply each term in $\frac{8 y + 9}{y + 8} = x$ by $y + 8$ .

$\frac{8 y + 9}{y + 8} (y + 8) = x (y + 8)$

Simplify the left side.

Cancel the common factor of $y + 8$ .

Cancel the common factor.

$\frac{8 y + 9}{y + 8} (y + 8) = x (y + 8)$

Rewrite the expression.

$8 y + 9 = x (y + 8)$

Simplify the right side.

Apply the distributive property.

$8 y + 9 = x y + x \cdot 8$

Move $8$ to the left of $x$ .

$8 y + 9 = x y + 8 x$

Solve the equation.

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

$8 y + 9 - x y = 8 x$

Subtract $9$ from both sides of the equation.

$8 y - x y = 8 x - 9$

Factor $y$ out of $8 y - x y$ .

Factor $y$ out of $8 y$ .

$y \cdot 8 - x y = 8 x - 9$

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

$y \cdot 8 + y (- x) = 8 x - 9$

Factor $y$ out of $y \cdot 8 + y (- x)$ .

$y (8 - x) = 8 x - 9$

Divide each term in $y (8 - x) = 8 x - 9$ by $8 - x$ and simplify.

Divide each term in $y (8 - x) = 8 x - 9$ by $8 - x$ .

$\frac{y (8 - x)}{8 - x} = \frac{8 x}{8 - x} + \frac{- 9}{8 - x}$

Simplify the left side.

Cancel the common factor of $8 - x$ .

Cancel the common factor.

$\frac{y (8 - x)}{8 - x} = \frac{8 x}{8 - x} + \frac{- 9}{8 - x}$

Divide $y$ by $1$ .

$y = \frac{8 x}{8 - x} + \frac{- 9}{8 - x}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{8 x - 9}{8 - x}$

Step 3

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

$f^{- 1} (x) = \frac{8 x - 9}{8 - x}$

Step 4

Verify if $f^{- 1} (x) = \frac{8 x - 9}{8 - x}$ is the inverse of $f (x) = \frac{8 x + 9}{x + 8}$ .

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 $f^{- 1} (\frac{8 x + 9}{x + 8})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{8 (\frac{8 x + 9}{x + 8}) - 9}{8 - (\frac{8 x + 9}{x + 8})}$

Multiply the numerator and denominator of the complex fraction by $x + 8$ .

Multiply $\frac{8 \frac{8 x + 9}{x + 8} - 9}{8 - \frac{8 x + 9}{x + 8}}$ by $\frac{x + 8}{x + 8}$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{x + 8}{x + 8} \cdot \frac{8 (\frac{8 x + 9}{x + 8}) - 9}{8 - \frac{8 x + 9}{x + 8}}$

Combine.

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (8 (\frac{8 x + 9}{x + 8}) - 9)}{(x + 8) (8 - \frac{8 x + 9}{x + 8})}$

Apply the distributive property.

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (8 (\frac{8 x + 9}{x + 8})) + (x + 8) \cdot - 9}{(x + 8) \cdot 8 + (x + 8) (- \frac{8 x + 9}{x + 8})}$

Cancel the common factor of $x + 8$ .

Move the leading negative in $- \frac{8 x + 9}{x + 8}$ into the numerator.

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (8 (\frac{8 x + 9}{x + 8})) + (x + 8) \cdot - 9}{(x + 8) \cdot 8 + (x + 8) (\frac{- (8 x + 9)}{x + 8})}$

Cancel the common factor.

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (8 (\frac{8 x + 9}{x + 8})) + (x + 8) \cdot - 9}{(x + 8) \cdot 8 + (x + 8) (\frac{- (8 x + 9)}{x + 8})}$

Rewrite the expression.

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (8 (\frac{8 x + 9}{x + 8})) + (x + 8) \cdot - 9}{(x + 8) \cdot 8 - (8 x + 9)}$

Simplify the numerator.

Factor $x + 8$ out of $(x + 8) \cdot 8 \frac{8 x + 9}{x + 8} + (x + 8) \cdot - 9$ .

Factor $x + 8$ out of $(x + 8) \cdot 8 \frac{8 x + 9}{x + 8}$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (8 (\frac{8 x + 9}{x + 8})) + (x + 8) \cdot - 9}{(x + 8) \cdot 8 - (8 x + 9)}$

Factor $x + 8$ out of $(x + 8) \cdot - 9$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (8 (\frac{8 x + 9}{x + 8})) + (x + 8) (- 9)}{(x + 8) \cdot 8 - (8 x + 9)}$

Factor $x + 8$ out of $(x + 8) (8 \frac{8 x + 9}{x + 8}) + (x + 8) (- 9)$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (8 (\frac{8 x + 9}{x + 8}) - 9)}{(x + 8) \cdot 8 - (8 x + 9)}$

Combine $8$ and $\frac{8 x + 9}{x + 8}$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{8 (8 x + 9)}{x + 8} - 9)}{(x + 8) \cdot 8 - (8 x + 9)}$

To write $- 9$ as a fraction with a common denominator, multiply by $\frac{x + 8}{x + 8}$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{8 (8 x + 9)}{x + 8} - 9 \cdot \frac{x + 8}{x + 8})}{(x + 8) \cdot 8 - (8 x + 9)}$

Combine $- 9$ and $\frac{x + 8}{x + 8}$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{8 (8 x + 9)}{x + 8} + \frac{- 9 (x + 8)}{x + 8})}{(x + 8) \cdot 8 - (8 x + 9)}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{8 (8 x + 9) - 9 (x + 8)}{x + 8})}{(x + 8) \cdot 8 - (8 x + 9)}$

Rewrite $\frac{8 (8 x + 9) - 9 (x + 8)}{x + 8}$ in a factored form.

Apply the distributive property.

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{8 (8 x) + 8 \cdot 9 - 9 (x + 8)}{x + 8})}{(x + 8) \cdot 8 - (8 x + 9)}$

Multiply $8$ by $8$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{64 x + 8 \cdot 9 - 9 (x + 8)}{x + 8})}{(x + 8) \cdot 8 - (8 x + 9)}$

Multiply $8$ by $9$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{64 x + 72 - 9 (x + 8)}{x + 8})}{(x + 8) \cdot 8 - (8 x + 9)}$

Apply the distributive property.

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{64 x + 72 - 9 x - 9 \cdot 8}{x + 8})}{(x + 8) \cdot 8 - (8 x + 9)}$

Multiply $- 9$ by $8$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{64 x + 72 - 9 x - 72}{x + 8})}{(x + 8) \cdot 8 - (8 x + 9)}$

Subtract $9 x$ from $64 x$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{55 x + 72 - 72}{x + 8})}{(x + 8) \cdot 8 - (8 x + 9)}$

Subtract $72$ from $72$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{55 x + 0}{x + 8})}{(x + 8) \cdot 8 - (8 x + 9)}$

Add $55 x$ and $0$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{55 x}{x + 8})}{(x + 8) \cdot 8 - (8 x + 9)}$

Simplify the denominator.

Apply the distributive property.

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{55 x}{x + 8})}{x \cdot 8 + 8 \cdot 8 - (8 x + 9)}$

Move $8$ to the left of $x$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{55 x}{x + 8})}{8 \cdot x + 8 \cdot 8 - (8 x + 9)}$

Multiply $8$ by $8$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{55 x}{x + 8})}{8 \cdot x + 64 - (8 x + 9)}$

Apply the distributive property.

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{55 x}{x + 8})}{8 x + 64 - (8 x) - 1 \cdot 9}$

Multiply $8$ by $- 1$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{55 x}{x + 8})}{8 x + 64 - 8 x - 1 \cdot 9}$

Multiply $- 1$ by $9$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{55 x}{x + 8})}{8 x + 64 - 8 x - 9}$

Subtract $8 x$ from $8 x$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{55 x}{x + 8})}{0 + 64 - 9}$

Add $0$ and $64$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{55 x}{x + 8})}{64 - 9}$

Subtract $9$ from $64$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{(x + 8) (\frac{55 x}{x + 8})}{55}$

Reduce the expression by cancelling the common factors.

Factor $\frac{55 x}{x + 8}$ out of $\frac{(x + 8) \frac{55 x}{x + 8}}{55}$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{55 x}{x + 8} \cdot \frac{x + 8}{55}$

Cancel the common factor of $55$ .

Factor $55$ out of $55 x$ .

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{55 (x)}{x + 8} \cdot \frac{x + 8}{55}$

Cancel the common factor.

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{55 x}{x + 8} \cdot \frac{x + 8}{55}$

Rewrite the expression.

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{x}{x + 8} \cdot (x + 8)$

Cancel the common factor of $x + 8$ .

Cancel the common factor.

$f^{- 1} (\frac{8 x + 9}{x + 8}) = \frac{x}{x + 8} \cdot (x + 8)$

Rewrite the expression.

$f^{- 1} (\frac{8 x + 9}{x + 8}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{8 x - 9}{8 - x})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{8 (\frac{8 x - 9}{8 - x}) + 9}{(\frac{8 x - 9}{8 - x}) + 8}$

Multiply the numerator and denominator of the complex fraction by $8 - x$ .

Multiply $\frac{8 \frac{8 x - 9}{8 - x} + 9}{\frac{8 x - 9}{8 - x} + 8}$ by $\frac{8 - x}{8 - x}$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{8 - x}{8 - x} \cdot \frac{8 (\frac{8 x - 9}{8 - x}) + 9}{\frac{8 x - 9}{8 - x} + 8}$

Combine.

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (8 (\frac{8 x - 9}{8 - x}) + 9)}{(8 - x) (\frac{8 x - 9}{8 - x} + 8)}$

Apply the distributive property.

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (8 (\frac{8 x - 9}{8 - x})) + (8 - x) \cdot 9}{(8 - x) (\frac{8 x - 9}{8 - x}) + (8 - x) \cdot 8}$

Cancel the common factor of $8 - x$ .

Cancel the common factor.

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (8 (\frac{8 x - 9}{8 - x})) + (8 - x) \cdot 9}{(8 - x) (\frac{8 x - 9}{8 - x}) + (8 - x) \cdot 8}$

Rewrite the expression.

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (8 (\frac{8 x - 9}{8 - x})) + (8 - x) \cdot 9}{8 x - 9 + (8 - x) \cdot 8}$

Simplify the numerator.

Factor $8 - x$ out of $(8 - x) \cdot 8 \frac{8 x - 9}{8 - x} + (8 - x) \cdot 9$ .

Factor $8 - x$ out of $(8 - x) \cdot 8 \frac{8 x - 9}{8 - x}$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (8 (\frac{8 x - 9}{8 - x})) + (8 - x) \cdot 9}{8 x - 9 + (8 - x) \cdot 8}$

Factor $8 - x$ out of $(8 - x) \cdot 9$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (8 (\frac{8 x - 9}{8 - x})) + (8 - x) (9)}{8 x - 9 + (8 - x) \cdot 8}$

Factor $8 - x$ out of $(8 - x) (8 \frac{8 x - 9}{8 - x}) + (8 - x) (9)$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (8 (\frac{8 x - 9}{8 - x}) + 9)}{8 x - 9 + (8 - x) \cdot 8}$

Combine $8$ and $\frac{8 x - 9}{8 - x}$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{8 (8 x - 9)}{8 - x} + 9)}{8 x - 9 + (8 - x) \cdot 8}$

To write $9$ as a fraction with a common denominator, multiply by $\frac{8 - x}{8 - x}$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{8 (8 x - 9)}{8 - x} + \frac{9 (8 - x)}{8 - x})}{8 x - 9 + (8 - x) \cdot 8}$

Combine the numerators over the common denominator.

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{8 (8 x - 9) + 9 (8 - x)}{8 - x})}{8 x - 9 + (8 - x) \cdot 8}$

Rewrite $\frac{8 (8 x - 9) + 9 (8 - x)}{8 - x}$ in a factored form.

Apply the distributive property.

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{8 (8 x) + 8 \cdot - 9 + 9 (8 - x)}{8 - x})}{8 x - 9 + (8 - x) \cdot 8}$

Multiply $8$ by $8$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{64 x + 8 \cdot - 9 + 9 (8 - x)}{8 - x})}{8 x - 9 + (8 - x) \cdot 8}$

Multiply $8$ by $- 9$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{64 x - 72 + 9 (8 - x)}{8 - x})}{8 x - 9 + (8 - x) \cdot 8}$

Apply the distributive property.

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{64 x - 72 + 9 \cdot 8 + 9 (- x)}{8 - x})}{8 x - 9 + (8 - x) \cdot 8}$

Multiply $9$ by $8$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{64 x - 72 + 72 + 9 (- x)}{8 - x})}{8 x - 9 + (8 - x) \cdot 8}$

Multiply $- 1$ by $9$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{64 x - 72 + 72 - 9 x}{8 - x})}{8 x - 9 + (8 - x) \cdot 8}$

Subtract $9 x$ from $64 x$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{55 x - 72 + 72}{8 - x})}{8 x - 9 + (8 - x) \cdot 8}$

Add $- 72$ and $72$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{55 x + 0}{8 - x})}{8 x - 9 + (8 - x) \cdot 8}$

Add $55 x$ and $0$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{55 x}{8 - x})}{8 x - 9 + (8 - x) \cdot 8}$

Simplify the denominator.

Apply the distributive property.

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{55 x}{8 - x})}{8 x - 9 + 8 \cdot 8 - x \cdot 8}$

Multiply $8$ by $8$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{55 x}{8 - x})}{8 x - 9 + 64 - x \cdot 8}$

Multiply $8$ by $- 1$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{55 x}{8 - x})}{8 x - 9 + 64 - 8 x}$

Subtract $8 x$ from $8 x$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{55 x}{8 - x})}{0 - 9 + 64}$

Subtract $9$ from $0$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{55 x}{8 - x})}{- 9 + 64}$

Add $- 9$ and $64$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{(8 - x) (\frac{55 x}{8 - x})}{55}$

Reduce the expression by cancelling the common factors.

Factor $\frac{55 x}{8 - x}$ out of $\frac{(8 - x) \frac{55 x}{8 - x}}{55}$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{55 x}{8 - x} \cdot \frac{8 - x}{55}$

Cancel the common factor of $55$ .

Factor $55$ out of $55 x$ .

$f (\frac{8 x - 9}{8 - x}) = \frac{55 (x)}{8 - x} \cdot \frac{8 - x}{55}$

Cancel the common factor.

$f (\frac{8 x - 9}{8 - x}) = \frac{55 x}{8 - x} \cdot \frac{8 - x}{55}$

Rewrite the expression.

$f (\frac{8 x - 9}{8 - x}) = \frac{x}{8 - x} \cdot (8 - x)$

Cancel the common factor of $8 - x$ .

Cancel the common factor.

$f (\frac{8 x - 9}{8 - x}) = \frac{x}{8 - x} \cdot (8 - x)$

Rewrite the expression.

$f (\frac{8 x - 9}{8 - x}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{8 x - 9}{8 - x}$ is the inverse of $f (x) = \frac{8 x + 9}{x + 8}$ .

$f^{- 1} (x) = \frac{8 x - 9}{8 - x}$

Find the Inverse (8x+9)/(x+8)

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