Find the Inverse (4x)/(x+5)

$\frac{4 x}{x + 5}$

Step 1

Interchange the variables.

$x = \frac{4 y}{y + 5}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{4 y}{y + 5} = x$ .

$\frac{4 y}{y + 5} = x$

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 $\frac{4 y}{y + 5} = x$ by $y + 5$ to eliminate the fractions.

Multiply each term in $\frac{4 y}{y + 5} = x$ by $y + 5$ .

$\frac{4 y}{y + 5} (y + 5) = x (y + 5)$

Simplify the left side.

Cancel the common factor of $y + 5$ .

Cancel the common factor.

$\frac{4 y}{y + 5} (y + 5) = x (y + 5)$

Rewrite the expression.

$4 y = x (y + 5)$

Simplify the right side.

Apply the distributive property.

$4 y = x y + x \cdot 5$

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

$4 y = x y + 5 x$

Solve the equation.

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

$4 y - x y = 5 x$

Factor $y$ out of $4 y - x y$ .

Factor $y$ out of $4 y$ .

$y \cdot 4 - x y = 5 x$

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

$y \cdot 4 + y (- x) = 5 x$

Factor $y$ out of $y \cdot 4 + y (- x)$ .

$y (4 - x) = 5 x$

Divide each term in $y (4 - x) = 5 x$ by $4 - x$ and simplify.

Divide each term in $y (4 - x) = 5 x$ by $4 - x$ .

$\frac{y (4 - x)}{4 - x} = \frac{5 x}{4 - x}$

Simplify the left side.

Cancel the common factor of $4 - x$ .

Cancel the common factor.

$\frac{y (4 - x)}{4 - x} = \frac{5 x}{4 - x}$

Divide $y$ by $1$ .

$y = \frac{5 x}{4 - x}$

Step 3

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

$f^{- 1} (x) = \frac{5 x}{4 - x}$

Step 4

Verify if $f^{- 1} (x) = \frac{5 x}{4 - x}$ is the inverse of $f (x) = \frac{4 x}{x + 5}$ .

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 $f^{- 1} (\frac{4 x}{x + 5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{5 (\frac{4 x}{x + 5})}{4 - (\frac{4 x}{x + 5})}$

Combine $5$ and $\frac{4 x}{x + 5}$ .

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{\frac{5 (4 x)}{x + 5}}{4 - \frac{4 x}{x + 5}}$

Simplify the denominator.

To write $4$ as a fraction with a common denominator, multiply by $\frac{x + 5}{x + 5}$ .

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{\frac{5 (4 x)}{x + 5}}{\frac{4 (x + 5)}{x + 5} - \frac{4 x}{x + 5}}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{\frac{5 (4 x)}{x + 5}}{\frac{4 (x + 5) - 4 x}{x + 5}}$

Reorder terms.

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{\frac{5 (4 x)}{x + 5}}{\frac{- 4 x + 4 (x + 5)}{x + 5}}$

Rewrite $\frac{- 4 x + 4 (x + 5)}{x + 5}$ in a factored form.

Factor $4$ out of $- 4 x + 4 (x + 5)$ .

Factor $4$ out of $- 4 x$ .

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{\frac{5 (4 x)}{x + 5}}{\frac{4 (- x) + 4 (x + 5)}{x + 5}}$

Factor $4$ out of $4 (- x) + 4 (x + 5)$ .

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{\frac{5 (4 x)}{x + 5}}{\frac{4 (- x + x + 5)}{x + 5}}$

Add $- x$ and $x$ .

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{\frac{5 (4 x)}{x + 5}}{\frac{4 (0 + 5)}{x + 5}}$

Add $0$ and $5$ .

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{\frac{5 (4 x)}{x + 5}}{\frac{4 \cdot 5}{x + 5}}$

Multiply $4$ by $5$ .

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{\frac{5 (4 x)}{x + 5}}{\frac{20}{x + 5}}$

Multiply $5$ by $4$ .

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{\frac{20 x}{x + 5}}{\frac{20}{x + 5}}$

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{20 x}{x + 5} \cdot \frac{x + 5}{20}$

Cancel the common factor of $20$ .

Factor $20$ out of $20 x$ .

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{20 (x)}{x + 5} \cdot \frac{x + 5}{20}$

Cancel the common factor.

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{20 x}{x + 5} \cdot \frac{x + 5}{20}$

Rewrite the expression.

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{x}{x + 5} \cdot (x + 5)$

Cancel the common factor of $x + 5$ .

Cancel the common factor.

$f^{- 1} (\frac{4 x}{x + 5}) = \frac{x}{x + 5} \cdot (x + 5)$

Rewrite the expression.

$f^{- 1} (\frac{4 x}{x + 5}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{5 x}{4 - x})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{5 x}{4 - x}) = \frac{4 (\frac{5 x}{4 - x})}{(\frac{5 x}{4 - x}) + 5}$

Combine $4$ and $\frac{5 x}{4 - x}$ .

$f (\frac{5 x}{4 - x}) = \frac{\frac{4 (5 x)}{4 - x}}{\frac{5 x}{4 - x} + 5}$

Simplify the denominator.

To write $5$ as a fraction with a common denominator, multiply by $\frac{4 - x}{4 - x}$ .

$f (\frac{5 x}{4 - x}) = \frac{\frac{4 (5 x)}{4 - x}}{\frac{5 x}{4 - x} + \frac{5 (4 - x)}{4 - x}}$

Combine the numerators over the common denominator.

$f (\frac{5 x}{4 - x}) = \frac{\frac{4 (5 x)}{4 - x}}{\frac{5 x + 5 (4 - x)}{4 - x}}$

Reorder terms.

$f (\frac{5 x}{4 - x}) = \frac{\frac{4 (5 x)}{4 - x}}{\frac{5 x + 5 (4 - x)}{- x + 4}}$

Rewrite $\frac{5 x + 5 (4 - x)}{- x + 4}$ in a factored form.

Factor $5$ out of $5 x + 5 (4 - x)$ .

Factor $5$ out of $5 x$ .

$f (\frac{5 x}{4 - x}) = \frac{\frac{4 (5 x)}{4 - x}}{\frac{5 (x) + 5 (4 - x)}{- x + 4}}$

Factor $5$ out of $5 (x) + 5 (4 - x)$ .

$f (\frac{5 x}{4 - x}) = \frac{\frac{4 (5 x)}{4 - x}}{\frac{5 (x + 4 - x)}{- x + 4}}$

Subtract $x$ from $x$ .

$f (\frac{5 x}{4 - x}) = \frac{\frac{4 (5 x)}{4 - x}}{\frac{5 (0 + 4)}{- x + 4}}$

Add $0$ and $4$ .

$f (\frac{5 x}{4 - x}) = \frac{\frac{4 (5 x)}{4 - x}}{\frac{5 \cdot 4}{- x + 4}}$

Multiply $5$ by $4$ .

$f (\frac{5 x}{4 - x}) = \frac{\frac{4 (5 x)}{4 - x}}{\frac{20}{- x + 4}}$

Multiply $4$ by $5$ .

$f (\frac{5 x}{4 - x}) = \frac{\frac{20 x}{4 - x}}{\frac{20}{- x + 4}}$

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{5 x}{4 - x}) = \frac{20 x}{4 - x} \cdot \frac{- x + 4}{20}$

Cancel the common factor of $20$ .

Factor $20$ out of $20 x$ .

$f (\frac{5 x}{4 - x}) = \frac{20 (x)}{4 - x} \cdot \frac{- x + 4}{20}$

Cancel the common factor.

$f (\frac{5 x}{4 - x}) = \frac{20 x}{4 - x} \cdot \frac{- x + 4}{20}$

Rewrite the expression.

$f (\frac{5 x}{4 - x}) = \frac{x}{4 - x} \cdot (- x + 4)$

Multiply $\frac{x}{4 - x}$ by $- x + 4$ .

$f (\frac{5 x}{4 - x}) = \frac{x (- x + 4)}{4 - x}$

Cancel the common factor of $- x + 4$ and $4 - x$ .

Reorder terms.

$f (\frac{5 x}{4 - x}) = \frac{x (- x + 4)}{- x + 4}$

Cancel the common factor.

$f (\frac{5 x}{4 - x}) = \frac{x (- x + 4)}{- x + 4}$

Divide $x$ by $1$ .

$f (\frac{5 x}{4 - x}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{5 x}{4 - x}$ is the inverse of $f (x) = \frac{4 x}{x + 5}$ .

$f^{- 1} (x) = \frac{5 x}{4 - x}$

Find the Inverse (4x)/(x+5)

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