Find the Inverse (5x)/(x+7)

$\frac{5 x}{x + 7}$

Step 1

Interchange the variables.

$x = \frac{5 y}{y + 7}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{5 y}{y + 7} = x$ .

$\frac{5 y}{y + 7} = x$

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.

$y + 7, 1$

Remove parentheses.

$y + 7, 1$

The LCM of one and any expression is the expression.

$y + 7$

Multiply each term in $\frac{5 y}{y + 7} = x$ by $y + 7$ to eliminate the fractions.

Multiply each term in $\frac{5 y}{y + 7} = x$ by $y + 7$ .

$\frac{5 y}{y + 7} (y + 7) = x (y + 7)$

Simplify the left side.

Cancel the common factor of $y + 7$ .

Cancel the common factor.

$\frac{5 y}{y + 7} (y + 7) = x (y + 7)$

Rewrite the expression.

$5 y = x (y + 7)$

Simplify the right side.

Apply the distributive property.

$5 y = x y + x \cdot 7$

Move $7$ to the left of $x$ .

$5 y = x y + 7 x$

Solve the equation.

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

$5 y - x y = 7 x$

Factor $y$ out of $5 y - x y$ .

Factor $y$ out of $5 y$ .

$y \cdot 5 - x y = 7 x$

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

$y \cdot 5 + y (- x) = 7 x$

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

$y (5 - x) = 7 x$

Divide each term in $y (5 - x) = 7 x$ by $5 - x$ and simplify.

Divide each term in $y (5 - x) = 7 x$ by $5 - x$ .

$\frac{y (5 - x)}{5 - x} = \frac{7 x}{5 - x}$

Simplify the left side.

Cancel the common factor of $5 - x$ .

Cancel the common factor.

$\frac{y (5 - x)}{5 - x} = \frac{7 x}{5 - x}$

Divide $y$ by $1$ .

$y = \frac{7 x}{5 - x}$

Step 3

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

$f^{- 1} (x) = \frac{7 x}{5 - x}$

Step 4

Verify if $f^{- 1} (x) = \frac{7 x}{5 - x}$ is the inverse of $f (x) = \frac{5 x}{x + 7}$ .

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 $f^{- 1} (\frac{5 x}{x + 7})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{7 (\frac{5 x}{x + 7})}{5 - (\frac{5 x}{x + 7})}$

Combine $7$ and $\frac{5 x}{x + 7}$ .

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{\frac{7 (5 x)}{x + 7}}{5 - \frac{5 x}{x + 7}}$

Simplify the denominator.

To write $5$ as a fraction with a common denominator, multiply by $\frac{x + 7}{x + 7}$ .

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{\frac{7 (5 x)}{x + 7}}{\frac{5 (x + 7)}{x + 7} - \frac{5 x}{x + 7}}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{\frac{7 (5 x)}{x + 7}}{\frac{5 (x + 7) - 5 x}{x + 7}}$

Reorder terms.

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{\frac{7 (5 x)}{x + 7}}{\frac{- 5 x + 5 (x + 7)}{x + 7}}$

Rewrite $\frac{- 5 x + 5 (x + 7)}{x + 7}$ in a factored form.

Factor $5$ out of $- 5 x + 5 (x + 7)$ .

Factor $5$ out of $- 5 x$ .

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{\frac{7 (5 x)}{x + 7}}{\frac{5 (- x) + 5 (x + 7)}{x + 7}}$

Factor $5$ out of $5 (- x) + 5 (x + 7)$ .

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{\frac{7 (5 x)}{x + 7}}{\frac{5 (- x + x + 7)}{x + 7}}$

Add $- x$ and $x$ .

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{\frac{7 (5 x)}{x + 7}}{\frac{5 (0 + 7)}{x + 7}}$

Add $0$ and $7$ .

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{\frac{7 (5 x)}{x + 7}}{\frac{5 \cdot 7}{x + 7}}$

Multiply $5$ by $7$ .

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{\frac{7 (5 x)}{x + 7}}{\frac{35}{x + 7}}$

Multiply $7$ by $5$ .

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{\frac{35 x}{x + 7}}{\frac{35}{x + 7}}$

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{35 x}{x + 7} \cdot \frac{x + 7}{35}$

Cancel the common factor of $35$ .

Factor $35$ out of $35 x$ .

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{35 (x)}{x + 7} \cdot \frac{x + 7}{35}$

Cancel the common factor.

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{35 x}{x + 7} \cdot \frac{x + 7}{35}$

Rewrite the expression.

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{x}{x + 7} \cdot (x + 7)$

Cancel the common factor of $x + 7$ .

Cancel the common factor.

$f^{- 1} (\frac{5 x}{x + 7}) = \frac{x}{x + 7} \cdot (x + 7)$

Rewrite the expression.

$f^{- 1} (\frac{5 x}{x + 7}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{7 x}{5 - x})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{7 x}{5 - x}) = \frac{5 (\frac{7 x}{5 - x})}{(\frac{7 x}{5 - x}) + 7}$

Combine $5$ and $\frac{7 x}{5 - x}$ .

$f (\frac{7 x}{5 - x}) = \frac{\frac{5 (7 x)}{5 - x}}{\frac{7 x}{5 - x} + 7}$

Simplify the denominator.

To write $7$ as a fraction with a common denominator, multiply by $\frac{5 - x}{5 - x}$ .

$f (\frac{7 x}{5 - x}) = \frac{\frac{5 (7 x)}{5 - x}}{\frac{7 x}{5 - x} + \frac{7 (5 - x)}{5 - x}}$

Combine the numerators over the common denominator.

$f (\frac{7 x}{5 - x}) = \frac{\frac{5 (7 x)}{5 - x}}{\frac{7 x + 7 (5 - x)}{5 - x}}$

Reorder terms.

$f (\frac{7 x}{5 - x}) = \frac{\frac{5 (7 x)}{5 - x}}{\frac{7 x + 7 (5 - x)}{- x + 5}}$

Rewrite $\frac{7 x + 7 (5 - x)}{- x + 5}$ in a factored form.

Factor $7$ out of $7 x + 7 (5 - x)$ .

Factor $7$ out of $7 x$ .

$f (\frac{7 x}{5 - x}) = \frac{\frac{5 (7 x)}{5 - x}}{\frac{7 (x) + 7 (5 - x)}{- x + 5}}$

Factor $7$ out of $7 (x) + 7 (5 - x)$ .

$f (\frac{7 x}{5 - x}) = \frac{\frac{5 (7 x)}{5 - x}}{\frac{7 (x + 5 - x)}{- x + 5}}$

Subtract $x$ from $x$ .

$f (\frac{7 x}{5 - x}) = \frac{\frac{5 (7 x)}{5 - x}}{\frac{7 (0 + 5)}{- x + 5}}$

Add $0$ and $5$ .

$f (\frac{7 x}{5 - x}) = \frac{\frac{5 (7 x)}{5 - x}}{\frac{7 \cdot 5}{- x + 5}}$

Multiply $7$ by $5$ .

$f (\frac{7 x}{5 - x}) = \frac{\frac{5 (7 x)}{5 - x}}{\frac{35}{- x + 5}}$

Multiply $5$ by $7$ .

$f (\frac{7 x}{5 - x}) = \frac{\frac{35 x}{5 - x}}{\frac{35}{- x + 5}}$

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{7 x}{5 - x}) = \frac{35 x}{5 - x} \cdot \frac{- x + 5}{35}$

Cancel the common factor of $35$ .

Factor $35$ out of $35 x$ .

$f (\frac{7 x}{5 - x}) = \frac{35 (x)}{5 - x} \cdot \frac{- x + 5}{35}$

Cancel the common factor.

$f (\frac{7 x}{5 - x}) = \frac{35 x}{5 - x} \cdot \frac{- x + 5}{35}$

Rewrite the expression.

$f (\frac{7 x}{5 - x}) = \frac{x}{5 - x} \cdot (- x + 5)$

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

$f (\frac{7 x}{5 - x}) = \frac{x (- x + 5)}{5 - x}$

Cancel the common factor of $- x + 5$ and $5 - x$ .

Reorder terms.

$f (\frac{7 x}{5 - x}) = \frac{x (- x + 5)}{- x + 5}$

Cancel the common factor.

$f (\frac{7 x}{5 - x}) = \frac{x (- x + 5)}{- x + 5}$

Divide $x$ by $1$ .

$f (\frac{7 x}{5 - x}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{7 x}{5 - x}$ is the inverse of $f (x) = \frac{5 x}{x + 7}$ .

$f^{- 1} (x) = \frac{7 x}{5 - x}$

Find the Inverse (5x)/(x+7)

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