Find the Inverse (3x-1)/(x+5)

$\frac{3 x - 1}{x + 5}$

Step 1

Interchange the variables.

$x = \frac{3 y - 1}{y + 5}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{3 y - 1}{y + 5} = x$ .

$\frac{3 y - 1}{y + 5} = x$

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 $\frac{3 y - 1}{y + 5} = x$ by $y + 5$ to eliminate the fractions.

Multiply each term in $\frac{3 y - 1}{y + 5} = x$ by $y + 5$ .

$\frac{3 y - 1}{y + 5} (y + 5) = x (y + 5)$

Simplify the left side.

Cancel the common factor of $y + 5$ .

Cancel the common factor.

$\frac{3 y - 1}{y + 5} (y + 5) = x (y + 5)$

Rewrite the expression.

$3 y - 1 = x (y + 5)$

Simplify the right side.

Apply the distributive property.

$3 y - 1 = x y + x \cdot 5$

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

$3 y - 1 = x y + 5 x$

Solve the equation.

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

$3 y - 1 - x y = 5 x$

Add $1$ to both sides of the equation.

$3 y - x y = 5 x + 1$

Factor $y$ out of $3 y - x y$ .

Factor $y$ out of $3 y$ .

$y \cdot 3 - x y = 5 x + 1$

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

$y \cdot 3 + y (- x) = 5 x + 1$

Factor $y$ out of $y \cdot 3 + y (- x)$ .

$y (3 - x) = 5 x + 1$

Divide each term in $y (3 - x) = 5 x + 1$ by $3 - x$ and simplify.

Divide each term in $y (3 - x) = 5 x + 1$ by $3 - x$ .

$\frac{y (3 - x)}{3 - x} = \frac{5 x}{3 - x} + \frac{1}{3 - x}$

Simplify the left side.

Cancel the common factor of $3 - x$ .

Cancel the common factor.

$\frac{y (3 - x)}{3 - x} = \frac{5 x}{3 - x} + \frac{1}{3 - x}$

Divide $y$ by $1$ .

$y = \frac{5 x}{3 - x} + \frac{1}{3 - x}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{5 x + 1}{3 - x}$

Step 3

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

$f^{- 1} (x) = \frac{5 x + 1}{3 - x}$

Step 4

Verify if $f^{- 1} (x) = \frac{5 x + 1}{3 - x}$ is the inverse of $f (x) = \frac{3 x - 1}{x + 5}$ .

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 $f^{- 1} (\frac{3 x - 1}{x + 5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{5 (\frac{3 x - 1}{x + 5}) + 1}{3 - (\frac{3 x - 1}{x + 5})}$

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

Multiply $\frac{5 \frac{3 x - 1}{x + 5} + 1}{3 - \frac{3 x - 1}{x + 5}}$ by $\frac{x + 5}{x + 5}$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{x + 5}{x + 5} \cdot \frac{5 (\frac{3 x - 1}{x + 5}) + 1}{3 - \frac{3 x - 1}{x + 5}}$

Combine.

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (5 (\frac{3 x - 1}{x + 5}) + 1)}{(x + 5) (3 - \frac{3 x - 1}{x + 5})}$

Apply the distributive property.

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (5 (\frac{3 x - 1}{x + 5})) + (x + 5) \cdot 1}{(x + 5) \cdot 3 + (x + 5) (- \frac{3 x - 1}{x + 5})}$

Cancel the common factor of $x + 5$ .

Move the leading negative in $- \frac{3 x - 1}{x + 5}$ into the numerator.

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (5 (\frac{3 x - 1}{x + 5})) + (x + 5) \cdot 1}{(x + 5) \cdot 3 + (x + 5) (\frac{- (3 x - 1)}{x + 5})}$

Cancel the common factor.

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (5 (\frac{3 x - 1}{x + 5})) + (x + 5) \cdot 1}{(x + 5) \cdot 3 + (x + 5) (\frac{- (3 x - 1)}{x + 5})}$

Rewrite the expression.

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (5 (\frac{3 x - 1}{x + 5})) + (x + 5) \cdot 1}{(x + 5) \cdot 3 - (3 x - 1)}$

Simplify the numerator.

Factor $x + 5$ out of $(x + 5) \cdot 5 \frac{3 x - 1}{x + 5} + (x + 5) \cdot 1$ .

Factor $x + 5$ out of $(x + 5) \cdot 5 \frac{3 x - 1}{x + 5}$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (5 (\frac{3 x - 1}{x + 5})) + (x + 5) \cdot 1}{(x + 5) \cdot 3 - (3 x - 1)}$

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

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (5 (\frac{3 x - 1}{x + 5})) + (x + 5) (1)}{(x + 5) \cdot 3 - (3 x - 1)}$

Factor $x + 5$ out of $(x + 5) (5 \frac{3 x - 1}{x + 5}) + (x + 5) (1)$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (5 (\frac{3 x - 1}{x + 5}) + 1)}{(x + 5) \cdot 3 - (3 x - 1)}$

Combine $5$ and $\frac{3 x - 1}{x + 5}$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{5 (3 x - 1)}{x + 5} + 1)}{(x + 5) \cdot 3 - (3 x - 1)}$

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

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{5 (3 x - 1)}{x + 5} + \frac{x + 5}{x + 5})}{(x + 5) \cdot 3 - (3 x - 1)}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{5 (3 x - 1) + x + 5}{x + 5})}{(x + 5) \cdot 3 - (3 x - 1)}$

Reorder terms.

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{x + 5 (3 x - 1) + 5}{x + 5})}{(x + 5) \cdot 3 - (3 x - 1)}$

Rewrite $\frac{x + 5 (3 x - 1) + 5}{x + 5}$ in a factored form.

Apply the distributive property.

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{x + 5 (3 x) + 5 \cdot - 1 + 5}{x + 5})}{(x + 5) \cdot 3 - (3 x - 1)}$

Multiply $3$ by $5$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{x + 15 x + 5 \cdot - 1 + 5}{x + 5})}{(x + 5) \cdot 3 - (3 x - 1)}$

Multiply $5$ by $- 1$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{x + 15 x - 5 + 5}{x + 5})}{(x + 5) \cdot 3 - (3 x - 1)}$

Add $x$ and $15 x$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{16 x - 5 + 5}{x + 5})}{(x + 5) \cdot 3 - (3 x - 1)}$

Add $- 5$ and $5$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{16 x + 0}{x + 5})}{(x + 5) \cdot 3 - (3 x - 1)}$

Add $16 x$ and $0$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{16 x}{x + 5})}{(x + 5) \cdot 3 - (3 x - 1)}$

Simplify the denominator.

Apply the distributive property.

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{16 x}{x + 5})}{x \cdot 3 + 5 \cdot 3 - (3 x - 1)}$

Move $3$ to the left of $x$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{16 x}{x + 5})}{3 \cdot x + 5 \cdot 3 - (3 x - 1)}$

Multiply $5$ by $3$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{16 x}{x + 5})}{3 \cdot x + 15 - (3 x - 1)}$

Apply the distributive property.

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{16 x}{x + 5})}{3 x + 15 - (3 x) + 1}$

Multiply $3$ by $- 1$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{16 x}{x + 5})}{3 x + 15 - 3 x + 1}$

Multiply $- 1$ by $- 1$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{16 x}{x + 5})}{3 x + 15 - 3 x + 1}$

Subtract $3 x$ from $3 x$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{16 x}{x + 5})}{0 + 15 + 1}$

Add $0$ and $15$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{16 x}{x + 5})}{15 + 1}$

Add $15$ and $1$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{(x + 5) (\frac{16 x}{x + 5})}{16}$

Reduce the expression by cancelling the common factors.

Factor $\frac{16 x}{x + 5}$ out of $\frac{(x + 5) \frac{16 x}{x + 5}}{16}$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{16 x}{x + 5} \cdot \frac{x + 5}{16}$

Cancel the common factor of $16$ .

Factor $16$ out of $16 x$ .

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{16 (x)}{x + 5} \cdot \frac{x + 5}{16}$

Cancel the common factor.

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{16 x}{x + 5} \cdot \frac{x + 5}{16}$

Rewrite the expression.

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{x}{x + 5} \cdot (x + 5)$

Cancel the common factor of $x + 5$ .

Cancel the common factor.

$f^{- 1} (\frac{3 x - 1}{x + 5}) = \frac{x}{x + 5} \cdot (x + 5)$

Rewrite the expression.

$f^{- 1} (\frac{3 x - 1}{x + 5}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{5 x + 1}{3 - x})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{3 (\frac{5 x + 1}{3 - x}) - 1}{(\frac{5 x + 1}{3 - x}) + 5}$

Multiply the numerator and denominator of the complex fraction by $3 - x$ .

Multiply $\frac{3 \frac{5 x + 1}{3 - x} - 1}{\frac{5 x + 1}{3 - x} + 5}$ by $\frac{3 - x}{3 - x}$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{3 - x}{3 - x} \cdot \frac{3 (\frac{5 x + 1}{3 - x}) - 1}{\frac{5 x + 1}{3 - x} + 5}$

Combine.

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (3 (\frac{5 x + 1}{3 - x}) - 1)}{(3 - x) (\frac{5 x + 1}{3 - x} + 5)}$

Apply the distributive property.

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (3 (\frac{5 x + 1}{3 - x})) + (3 - x) \cdot - 1}{(3 - x) (\frac{5 x + 1}{3 - x}) + (3 - x) \cdot 5}$

Cancel the common factor of $3 - x$ .

Cancel the common factor.

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (3 (\frac{5 x + 1}{3 - x})) + (3 - x) \cdot - 1}{(3 - x) (\frac{5 x + 1}{3 - x}) + (3 - x) \cdot 5}$

Rewrite the expression.

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (3 (\frac{5 x + 1}{3 - x})) + (3 - x) \cdot - 1}{5 x + 1 + (3 - x) \cdot 5}$

Simplify the numerator.

Factor $3 - x$ out of $(3 - x) \cdot 3 \frac{5 x + 1}{3 - x} + (3 - x) \cdot - 1$ .

Factor $3 - x$ out of $(3 - x) \cdot 3 \frac{5 x + 1}{3 - x}$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (3 (\frac{5 x + 1}{3 - x})) + (3 - x) \cdot - 1}{5 x + 1 + (3 - x) \cdot 5}$

Factor $3 - x$ out of $(3 - x) \cdot - 1$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (3 (\frac{5 x + 1}{3 - x})) + (3 - x) (- 1)}{5 x + 1 + (3 - x) \cdot 5}$

Factor $3 - x$ out of $(3 - x) (3 \frac{5 x + 1}{3 - x}) + (3 - x) (- 1)$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (3 (\frac{5 x + 1}{3 - x}) - 1)}{5 x + 1 + (3 - x) \cdot 5}$

Combine $3$ and $\frac{5 x + 1}{3 - x}$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{3 (5 x + 1)}{3 - x} - 1)}{5 x + 1 + (3 - x) \cdot 5}$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3 - x}{3 - x}$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{3 (5 x + 1)}{3 - x} - 1 \cdot \frac{3 - x}{3 - x})}{5 x + 1 + (3 - x) \cdot 5}$

Combine $- 1$ and $\frac{3 - x}{3 - x}$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{3 (5 x + 1)}{3 - x} + \frac{- (3 - x)}{3 - x})}{5 x + 1 + (3 - x) \cdot 5}$

Combine the numerators over the common denominator.

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{3 (5 x + 1) - (3 - x)}{3 - x})}{5 x + 1 + (3 - x) \cdot 5}$

Reorder terms.

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{3 (5 x + 1) - (3 - x)}{- x + 3})}{5 x + 1 + (3 - x) \cdot 5}$

Rewrite $\frac{3 (5 x + 1) - (3 - x)}{- x + 3}$ in a factored form.

Apply the distributive property.

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{3 (5 x) + 3 \cdot 1 - (3 - x)}{- x + 3})}{5 x + 1 + (3 - x) \cdot 5}$

Multiply $5$ by $3$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{15 x + 3 \cdot 1 - (3 - x)}{- x + 3})}{5 x + 1 + (3 - x) \cdot 5}$

Multiply $3$ by $1$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{15 x + 3 - (3 - x)}{- x + 3})}{5 x + 1 + (3 - x) \cdot 5}$

Apply the distributive property.

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{15 x + 3 - 1 \cdot 3 + x}{- x + 3})}{5 x + 1 + (3 - x) \cdot 5}$

Multiply $- 1$ by $3$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{15 x + 3 - 3 + x}{- x + 3})}{5 x + 1 + (3 - x) \cdot 5}$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{15 x + 3 - 3 + 1 x}{- x + 3})}{5 x + 1 + (3 - x) \cdot 5}$

Multiply $x$ by $1$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{15 x + 3 - 3 + x}{- x + 3})}{5 x + 1 + (3 - x) \cdot 5}$

Add $15 x$ and $x$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{16 x + 3 - 3}{- x + 3})}{5 x + 1 + (3 - x) \cdot 5}$

Subtract $3$ from $3$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{16 x + 0}{- x + 3})}{5 x + 1 + (3 - x) \cdot 5}$

Add $16 x$ and $0$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{16 x}{- x + 3})}{5 x + 1 + (3 - x) \cdot 5}$

Simplify the denominator.

Apply the distributive property.

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{16 x}{- x + 3})}{5 x + 1 + 3 \cdot 5 - x \cdot 5}$

Multiply $3$ by $5$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{16 x}{- x + 3})}{5 x + 1 + 15 - x \cdot 5}$

Multiply $5$ by $- 1$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{16 x}{- x + 3})}{5 x + 1 + 15 - 5 x}$

Subtract $5 x$ from $5 x$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{16 x}{- x + 3})}{0 + 1 + 15}$

Add $0$ and $1$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{16 x}{- x + 3})}{1 + 15}$

Add $1$ and $15$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{(3 - x) (\frac{16 x}{- x + 3})}{16}$

Simplify terms.

Factor $\frac{16 x}{- x + 3}$ out of $\frac{(3 - x) \frac{16 x}{- x + 3}}{16}$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{16 x}{- x + 3} \cdot \frac{3 - x}{16}$

Cancel the common factor of $16$ .

Factor $16$ out of $16 x$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{16 (x)}{- x + 3} \cdot \frac{3 - x}{16}$

Cancel the common factor.

$f (\frac{5 x + 1}{3 - x}) = \frac{16 x}{- x + 3} \cdot \frac{3 - x}{16}$

Rewrite the expression.

$f (\frac{5 x + 1}{3 - x}) = \frac{x}{- x + 3} \cdot (3 - x)$

Multiply $\frac{x}{- x + 3}$ by $3 - x$ .

$f (\frac{5 x + 1}{3 - x}) = \frac{x (3 - x)}{- x + 3}$

Cancel the common factor of $3 - x$ and $- x + 3$ .

Reorder terms.

$f (\frac{5 x + 1}{3 - x}) = \frac{x (- x + 3)}{- x + 3}$

Cancel the common factor.

$f (\frac{5 x + 1}{3 - x}) = \frac{x (- x + 3)}{- x + 3}$

Divide $x$ by $1$ .

$f (\frac{5 x + 1}{3 - x}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{5 x + 1}{3 - x}$ is the inverse of $f (x) = \frac{3 x - 1}{x + 5}$ .

$f^{- 1} (x) = \frac{5 x + 1}{3 - x}$

Find the Inverse (3x-1)/(x+5)

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