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

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

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.

$y - 1, 1$

Remove parentheses.

$y - 1, 1$

The LCM of one and any expression is the expression.

$y - 1$

Multiply each term in $\frac{5 y - 3}{y - 1} = x$ by $y - 1$ to eliminate the fractions.

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

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

Simplify the left side.

Cancel the common factor of $y - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Simplify by multiplying through.

Apply the distributive property.

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

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

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

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

$5 y - 3 = x y - x$

Solve the equation.

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

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

Add $3$ to both sides of the equation.

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

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

Factor $y$ out of $5 y$ .

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

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

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

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

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

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

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

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

Simplify the left side.

Cancel the common factor of $5 - x$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Combine the numerators over the common denominator.

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

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

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

Rewrite $3$ as $- 1 (- 3)$ .

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

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

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

Simplify the expression.

Rewrite $- (x - 3)$ as $- 1 (x - 3)$ .

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

Move the negative in front of the fraction.

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

Step 3

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

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

Step 4

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

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

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

Multiply the numerator and denominator of the complex fraction by $x - 1$ .

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

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

Combine.

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

Apply the distributive property.

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

Simplify by cancelling.

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 1$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Apply the distributive property.

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

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

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

Multiply $- 1$ by $- 3$ .

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

Subtract $3 x$ from $5 x$ .

$f^{- 1} (\frac{5 x - 3}{x - 1}) = - \frac{2 x - 3 + 3}{(x - 1) \cdot 5 - (5 x - 3)}$

Add $- 3$ and $3$ .

$f^{- 1} (\frac{5 x - 3}{x - 1}) = - \frac{2 x + 0}{(x - 1) \cdot 5 - (5 x - 3)}$

Add $2 x$ and $0$ .

$f^{- 1} (\frac{5 x - 3}{x - 1}) = - \frac{2 x}{(x - 1) \cdot 5 - (5 x - 3)}$

Simplify the denominator.

Apply the distributive property.

$f^{- 1} (\frac{5 x - 3}{x - 1}) = - \frac{2 x}{x \cdot 5 - 1 \cdot 5 - (5 x - 3)}$

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

$f^{- 1} (\frac{5 x - 3}{x - 1}) = - \frac{2 x}{5 \cdot x - 1 \cdot 5 - (5 x - 3)}$

Multiply $- 1$ by $5$ .

$f^{- 1} (\frac{5 x - 3}{x - 1}) = - \frac{2 x}{5 \cdot x - 5 - (5 x - 3)}$

Apply the distributive property.

$f^{- 1} (\frac{5 x - 3}{x - 1}) = - \frac{2 x}{5 x - 5 - (5 x) + 3}$

Multiply $5$ by $- 1$ .

$f^{- 1} (\frac{5 x - 3}{x - 1}) = - \frac{2 x}{5 x - 5 - 5 x + 3}$

Multiply $- 1$ by $- 3$ .

$f^{- 1} (\frac{5 x - 3}{x - 1}) = - \frac{2 x}{5 x - 5 - 5 x + 3}$

Subtract $5 x$ from $5 x$ .

$f^{- 1} (\frac{5 x - 3}{x - 1}) = - \frac{2 x}{0 - 5 + 3}$

Subtract $5$ from $0$ .

$f^{- 1} (\frac{5 x - 3}{x - 1}) = - \frac{2 x}{- 5 + 3}$

Add $- 5$ and $3$ .

$f^{- 1} (\frac{5 x - 3}{x - 1}) = - \frac{2 x}{- 2}$

Reduce the expression by cancelling the common factors.

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

Factor $2$ out of $2 x$ .

$f^{- 1} (\frac{5 x - 3}{x - 1}) = - \frac{2 (x)}{- 2}$

Move the negative one from the denominator of $\frac{x}{- 1}$ .

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

Rewrite $- 1 \cdot x$ as $- x$ .

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

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

Set up the composite result function.

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

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

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

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

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

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

Combine.

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

Apply the distributive property.

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

Cancel the common factor of $5 - x$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

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

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

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

Factor $5 - x$ out of $(5 - x) \cdot - 3$ .

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

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

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

Multiply $5 (- \frac{x - 3}{5 - x})$ .

Multiply $- 1$ by $5$ .

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

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

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

Move the negative in front of the fraction.

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

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{5 - x}{5 - x}$ .

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

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

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

Combine the numerators over the common denominator.

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

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

Factor $- 1$ out of $- 5 (x - 3) - 3 (5 - x)$ .

Reorder $5$ and $- x$ .

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

Factor $- 1$ out of $- 5 (x - 3)$ .

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

Factor $- 1$ out of $- 3 (- x + 5)$ .

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

Factor $- 1$ out of $- (5 (x - 3)) - (3 (- x + 5))$ .

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

Apply the distributive property.

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

Multiply $5$ by $- 3$ .

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

Apply the distributive property.

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

Multiply $- 1$ by $3$ .

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

Multiply $3$ by $5$ .

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

Subtract $3 x$ from $5 x$ .

$f (- \frac{x - 3}{5 - x}) = \frac{(5 - x) (\frac{- (2 x - 15 + 15)}{5 - x})}{- (x - 3) + (5 - x) \cdot - 1}$

Add $- 15$ and $15$ .

$f (- \frac{x - 3}{5 - x}) = \frac{(5 - x) (\frac{- (2 x + 0)}{5 - x})}{- (x - 3) + (5 - x) \cdot - 1}$

Add $2 x$ and $0$ .

$f (- \frac{x - 3}{5 - x}) = \frac{(5 - x) (\frac{- 1 \cdot (2 x)}{5 - x})}{- (x - 3) + (5 - x) \cdot - 1}$

Multiply $- 1$ by $2$ .

$f (- \frac{x - 3}{5 - x}) = \frac{(5 - x) (\frac{- 2 x}{5 - x})}{- (x - 3) + (5 - x) \cdot - 1}$

Move the negative in front of the fraction.

$f (- \frac{x - 3}{5 - x}) = \frac{(5 - x) (- \frac{(2) x}{5 - x})}{- (x - 3) + (5 - x) \cdot - 1}$

Remove unnecessary parentheses.

$f (- \frac{x - 3}{5 - x}) = \frac{(5 - x) \cdot (- 1 \frac{2 x}{5 - x})}{- (x - 3) + (5 - x) \cdot - 1}$

Factor out negative.

$f (- \frac{x - 3}{5 - x}) = \frac{- (5 - x) \frac{2 x}{5 - x}}{- (x - 3) + (5 - x) \cdot - 1}$

Simplify the denominator.

Apply the distributive property.

$f (- \frac{x - 3}{5 - x}) = \frac{- (5 - x) \frac{2 x}{5 - x}}{- x + 3 + (5 - x) \cdot - 1}$

Multiply $- 1$ by $- 3$ .

$f (- \frac{x - 3}{5 - x}) = \frac{- (5 - x) \frac{2 x}{5 - x}}{- x + 3 + (5 - x) \cdot - 1}$

Apply the distributive property.

$f (- \frac{x - 3}{5 - x}) = \frac{- (5 - x) \frac{2 x}{5 - x}}{- x + 3 + 5 \cdot - 1 - x \cdot - 1}$

Multiply $5$ by $- 1$ .

$f (- \frac{x - 3}{5 - x}) = \frac{- (5 - x) \frac{2 x}{5 - x}}{- x + 3 - 5 - x \cdot - 1}$

Multiply $- x \cdot - 1$ .

Multiply $- 1$ by $- 1$ .

$f (- \frac{x - 3}{5 - x}) = \frac{- (5 - x) \frac{2 x}{5 - x}}{- x + 3 - 5 + 1 x}$

Multiply $x$ by $1$ .

$f (- \frac{x - 3}{5 - x}) = \frac{- (5 - x) \frac{2 x}{5 - x}}{- x + 3 - 5 + x}$

Add $- x$ and $x$ .

$f (- \frac{x - 3}{5 - x}) = \frac{- (5 - x) \frac{2 x}{5 - x}}{0 + 3 - 5}$

Add $0$ and $3$ .

$f (- \frac{x - 3}{5 - x}) = \frac{- (5 - x) \frac{2 x}{5 - x}}{3 - 5}$

Subtract $5$ from $3$ .

$f (- \frac{x - 3}{5 - x}) = \frac{- (5 - x) \frac{2 x}{5 - x}}{- 2}$

Simplify terms.

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

$f (- \frac{x - 3}{5 - x}) = \frac{2 x}{5 - x} \cdot \frac{- (5 - x)}{- 2}$

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

$f (- \frac{x - 3}{5 - x}) = \frac{2 (x)}{5 - x} \cdot \frac{- (5 - x)}{- 2}$

Factor $2$ out of $- 2$ .

$f (- \frac{x - 3}{5 - x}) = \frac{2 (x)}{5 - x} \cdot \frac{- (5 - x)}{2 (- 1)}$

Cancel the common factor.

$f (- \frac{x - 3}{5 - x}) = \frac{2 x}{5 - x} \cdot \frac{- (5 - x)}{2 \cdot - 1}$

Rewrite the expression.

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

Cancel the common factor of $5 - x$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Simplify the expression.

Move the negative one from the denominator of $\frac{x \cdot - 1}{- 1}$ .

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

Rewrite $- 1 \cdot (x \cdot - 1)$ as $- (x \cdot - 1)$ .

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

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

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

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

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

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

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

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

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