Find the Inverse y=(2x+3)/(x-1)

$y = \frac{2 x + 3}{x - 1}$

Step 1

Interchange the variables.

$x = \frac{2 y + 3}{y - 1}$

Step 2

Solve for $y$ .

Multiply the equation by $y - 1$ .

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

Simplify the left side.

Simplify $(y - 1) (x)$ .

Apply the distributive property.

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

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

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

Simplify the right side.

Cancel the common factor of $y - 1$ .

Cancel the common factor.

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

Rewrite the expression.

$y x - x = 2 y + 3$

Solve for $y$ .

Subtract $2 y$ from both sides of the equation.

$y x - x - 2 y = 3$

Add $x$ to both sides of the equation.

$y x - 2 y = 3 + x$

Factor $y$ out of $y x - 2 y$ .

Factor $y$ out of $y x$ .

$y (x) - 2 y = 3 + x$

Factor $y$ out of $- 2 y$ .

$y (x) + y \cdot - 2 = 3 + x$

Factor $y$ out of $y (x) + y \cdot - 2$ .

$y (x - 2) = 3 + x$

Divide each term in $y (x - 2) = 3 + x$ by $x - 2$ and simplify.

Divide each term in $y (x - 2) = 3 + x$ by $x - 2$ .

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

Simplify the left side.

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

$y = \frac{3}{x - 2} + \frac{x}{x - 2}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{3 + x}{x - 2}$

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = \frac{3 + x}{x - 2}$ is the inverse of $f (x) = \frac{2 x + 3}{x - 1}$ .

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 $f^{- 1} (\frac{2 x + 3}{x - 1})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Remove parentheses.

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

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

Multiply $\frac{3 + \frac{2 x + 3}{x - 1}}{\frac{2 x + 3}{x - 1} - 2}$ by $\frac{x - 1}{x - 1}$ .

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

Combine.

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

Apply the distributive property.

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

Simplify by cancelling.

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Apply the distributive property.

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

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

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

Multiply $- 1$ by $3$ .

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

Add $3 x$ and $2 x$ .

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

Add $- 3$ and $3$ .

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

Add $5 x$ and $0$ .

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

Simplify the denominator.

Apply the distributive property.

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

Move $- 2$ to the left of $x$ .

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

Multiply $- 1$ by $- 2$ .

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

Subtract $2 x$ from $2 x$ .

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

Add $0$ and $3$ .

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

Add $3$ and $2$ .

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

Evaluate $f (\frac{3 + x}{x - 2})$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Multiply the numerator and denominator of the complex fraction by $x - 2$ .

Multiply $\frac{2 \frac{3 + x}{x - 2} + 3}{\frac{3 + x}{x - 2} - 1}$ by $\frac{x - 2}{x - 2}$ .

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

Combine.

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

Apply the distributive property.

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

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Factor $x - 2$ out of $(x - 2) \cdot 2 \frac{3 + x}{x - 2} + (x - 2) \cdot 3$ .

Factor $x - 2$ out of $(x - 2) \cdot 2 \frac{3 + x}{x - 2}$ .

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

Factor $x - 2$ out of $(x - 2) \cdot 3$ .

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

Factor $x - 2$ out of $(x - 2) (2 \frac{3 + x}{x - 2}) + (x - 2) (3)$ .

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

Combine $2$ and $\frac{3 + x}{x - 2}$ .

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

To write $3$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

Rewrite $\frac{3 (x - 2) + 2 (3 + x)}{x - 2}$ in a factored form.

Apply the distributive property.

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

Multiply $3$ by $- 2$ .

$f (\frac{3 + x}{x - 2}) = \frac{(x - 2) (\frac{3 x - 6 + 2 (3 + x)}{x - 2})}{3 + x + (x - 2) \cdot - 1}$

Apply the distributive property.

$f (\frac{3 + x}{x - 2}) = \frac{(x - 2) (\frac{3 x - 6 + 2 \cdot 3 + 2 x}{x - 2})}{3 + x + (x - 2) \cdot - 1}$

Multiply $2$ by $3$ .

$f (\frac{3 + x}{x - 2}) = \frac{(x - 2) (\frac{3 x - 6 + 6 + 2 x}{x - 2})}{3 + x + (x - 2) \cdot - 1}$

Add $3 x$ and $2 x$ .

$f (\frac{3 + x}{x - 2}) = \frac{(x - 2) (\frac{5 x - 6 + 6}{x - 2})}{3 + x + (x - 2) \cdot - 1}$

Add $- 6$ and $6$ .

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

Add $5 x$ and $0$ .

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

Simplify the denominator.

Apply the distributive property.

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

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

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

Multiply $- 2$ by $- 1$ .

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

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

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

Add $3$ and $2$ .

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

Subtract $x$ from $x$ .

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

Add $0$ and $5$ .

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

Reduce the expression by cancelling the common factors.

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

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

Cancel the common factor of $5$ .

Factor $5$ out of $5 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{3 + x}{x - 2}$ is the inverse of $f (x) = \frac{2 x + 3}{x - 1}$ .

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

Find the Inverse y=(2x+3)/(x-1)

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