Find the Inverse (3x)/(2x-1)

$\frac{3 x}{2 x - 1}$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Multiply the equation by $2 y - 1$ .

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

Simplify the left side.

Simplify $(2 y - 1) (x)$ .

Apply the distributive property.

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

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

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

Simplify the right side.

Cancel the common factor of $2 y - 1$ .

Cancel the common factor.

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

Rewrite the expression.

$2 y x - x = 3 y$

Solve for $y$ .

Subtract $3 y$ from both sides of the equation.

$2 y x - x - 3 y = 0$

Add $x$ to both sides of the equation.

$2 y x - 3 y = x$

Factor $y$ out of $2 y x - 3 y$ .

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

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

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

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

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

$y (2 x - 3) = x$

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

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

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

Simplify the left side.

Cancel the common factor of $2 x - 3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Step 3

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

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

Step 4

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

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

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

Multiply the numerator by the reciprocal of the denominator.

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

Simplify the denominator.

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

Combine $2$ and $\frac{3 x}{2 x - 1}$ .

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

Multiply $3$ by $2$ .

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

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{2 x - 1}{2 x - 1}$ .

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

Combine $- 3$ and $\frac{2 x - 1}{2 x - 1}$ .

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

Combine the numerators over the common denominator.

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

Rewrite $\frac{6 x - 3 (2 x - 1)}{2 x - 1}$ in a factored form.

Factor $3$ out of $6 x - 3 (2 x - 1)$ .

Factor $3$ out of $6 x$ .

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

Factor $3$ out of $- 3 (2 x - 1)$ .

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

Factor $3$ out of $3 (2 x) + 3 (- (2 x - 1))$ .

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

Apply the distributive property.

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

Multiply $2$ by $- 1$ .

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

Multiply $- 1$ by $- 1$ .

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

Subtract $2 x$ from $2 x$ .

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

Add $0$ and $1$ .

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

Multiply $3$ by $1$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

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

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

Cancel the common factor of $3$ .

Factor $3$ out of $3 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2 x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Set up the composite result function.

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

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

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

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

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

Simplify the denominator.

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

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

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2 x - 3}{2 x - 3}$ .

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

Combine $- 1$ and $\frac{2 x - 3}{2 x - 3}$ .

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

Combine the numerators over the common denominator.

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

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

Apply the distributive property.

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

Multiply $2$ by $- 1$ .

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

Multiply $- 1$ by $- 3$ .

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

Subtract $2 x$ from $2 x$ .

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

Add $0$ and $3$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $3$ .

Factor $3$ out of $3 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2 x - 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Find the Inverse (3x)/(2x-1)

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