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

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{2 y + 1}{3 - 2 y} = x$ .

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

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.

$3 - 2 y, 1$

Remove parentheses.

$3 - 2 y, 1$

The LCM of one and any expression is the expression.

$3 - 2 y$

Multiply each term in $\frac{2 y + 1}{3 - 2 y} = x$ by $3 - 2 y$ to eliminate the fractions.

Multiply each term in $\frac{2 y + 1}{3 - 2 y} = x$ by $3 - 2 y$ .

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

Simplify the left side.

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

Cancel the common factor.

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

Rewrite the expression.

$2 y + 1 = x (3 - 2 y)$

Simplify the right side.

Apply the distributive property.

$2 y + 1 = x \cdot 3 + x (- 2 y)$

Reorder.

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

$2 y + 1 = 3 \cdot x + x (- 2 y)$

Rewrite using the commutative property of multiplication.

$2 y + 1 = 3 x - 2 x y$

Solve the equation.

Add $2 x y$ to both sides of the equation.

$2 y + 1 + 2 x y = 3 x$

Subtract $1$ from both sides of the equation.

$2 y + 2 x y = 3 x - 1$

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

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

$2 y (1) + 2 x y = 3 x - 1$

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

$2 y (1) + 2 y (x) = 3 x - 1$

Factor $2 y$ out of $2 y (1) + 2 y (x)$ .

$2 y (1 + x) = 3 x - 1$

Divide each term in $2 y (1 + x) = 3 x - 1$ by $2 (1 + x)$ and simplify.

Divide each term in $2 y (1 + x) = 3 x - 1$ by $2 (1 + x)$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $1 + x$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 3

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

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

Step 4

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

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

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

Remove parentheses.

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

Combine the numerators over the common denominator.

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

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

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

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

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

Simplify terms.

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $2$ by $3$ .

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

Multiply $3$ by $1$ .

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

Apply the distributive property.

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

Multiply $- 1$ by $3$ .

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

Multiply $- 2$ by $- 1$ .

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

Add $6 x$ and $2 x$ .

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{\frac{8 x + 3 - 3}{3 - 2 x}}{2 (1 + \frac{2 x + 1}{3 - 2 x})}$

Subtract $3$ from $3$ .

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{\frac{8 x + 0}{3 - 2 x}}{2 (1 + \frac{2 x + 1}{3 - 2 x})}$

Add $8 x$ and $0$ .

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{\frac{8 x}{3 - 2 x}}{2 (1 + \frac{2 x + 1}{3 - 2 x})}$

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{8 x}{3 - 2 x} \cdot \frac{1}{2 (1 + \frac{2 x + 1}{3 - 2 x})}$

Simplify terms.

Cancel the common factor of $2$ .

Factor $2$ out of $8 x$ .

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{2 (4 x)}{3 - 2 x} \cdot \frac{1}{2 (1 + \frac{2 x + 1}{3 - 2 x})}$

Cancel the common factor.

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{2 (4 x)}{3 - 2 x} \cdot \frac{1}{2 (1 + \frac{2 x + 1}{3 - 2 x})}$

Rewrite the expression.

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{4 x}{3 - 2 x} \cdot \frac{1}{1 + \frac{2 x + 1}{3 - 2 x}}$

Multiply $\frac{4 x}{3 - 2 x}$ by $\frac{1}{1 + \frac{2 x + 1}{3 - 2 x}}$ .

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{4 x}{(3 - 2 x) (1 + \frac{2 x + 1}{3 - 2 x})}$

Simplify the denominator.

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

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{4 x}{(3 - 2 x) (\frac{3 - 2 x}{3 - 2 x} + \frac{2 x + 1}{3 - 2 x})}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{4 x}{(3 - 2 x) (\frac{3 - 2 x + 2 x + 1}{3 - 2 x})}$

Reorder terms.

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{4 x}{(3 - 2 x) (\frac{2 x - 2 x + 3 + 1}{- 2 x + 3})}$

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

Subtract $2 x$ from $2 x$ .

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{4 x}{(3 - 2 x) (\frac{0 + 3 + 1}{- 2 x + 3})}$

Add $0$ and $3$ .

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{4 x}{(3 - 2 x) (\frac{3 + 1}{- 2 x + 3})}$

Add $3$ and $1$ .

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{4 x}{(3 - 2 x) (\frac{4}{- 2 x + 3})}$

Factor $\frac{4}{- 2 x + 3}$ out of $\frac{4 x}{(3 - 2 x) \frac{4}{- 2 x + 3}}$ .

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{- 2 x + 3}{4} \cdot \frac{4 x}{3 - 2 x}$

Cancel the common factor of $4$ .

Factor $4$ out of $4 x$ .

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{- 2 x + 3}{4} \cdot \frac{4 (x)}{3 - 2 x}$

Cancel the common factor.

$f^{- 1} (\frac{2 x + 1}{3 - 2 x}) = \frac{- 2 x + 3}{4} \cdot \frac{4 x}{3 - 2 x}$

Rewrite the expression.

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

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

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

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

Reorder terms.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Apply the distributive property.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2 (1 + x)}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Combine the numerators over the common denominator.

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

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

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

Add $3 x$ and $x$ .

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x - 1 + 1}{x + 1}}{3 - 2 (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)})}$

Add $- 1$ and $1$ .

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x + 0}{x + 1}}{3 - 2 (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)})}$

Add $4 x$ and $0$ .

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{3 - 2 (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)})}$

Simplify the denominator.

Apply the distributive property.

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{3 - 2 \frac{3 x}{2 (1 + x)} - 2 (- \frac{1}{2 (1 + x)})}$

Cancel the common factor of $2$ .

Factor $2$ out of $- 2$ .

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{3 + 2 (- 1) (\frac{3 x}{2 (1 + x)}) - 2 (- \frac{1}{2 (1 + x)})}$

Cancel the common factor.

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{3 + 2 \cdot (- 1 \frac{3 x}{2 (1 + x)}) - 2 (- \frac{1}{2 (1 + x)})}$

Rewrite the expression.

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{3 - 1 \frac{3 x}{1 + x} - 2 (- \frac{1}{2 (1 + x)})}$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2 (1 + x)}$ into the numerator.

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{3 - 1 \frac{3 x}{1 + x} - 2 \frac{- 1}{2 (1 + x)}}$

Factor $2$ out of $- 2$ .

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{3 - 1 \frac{3 x}{1 + x} + 2 (- 1) (\frac{- 1}{2 (1 + x)})}$

Cancel the common factor.

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{3 - 1 \frac{3 x}{1 + x} + 2 \cdot (- 1 \frac{- 1}{2 (1 + x)})}$

Rewrite the expression.

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{3 - 1 \frac{3 x}{1 + x} - 1 \frac{- 1}{1 + x}}$

Simplify each term.

Rewrite $- 1 \frac{3 x}{1 + x}$ as $- \frac{3 x}{1 + x}$ .

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{3 - \frac{3 x}{1 + x} - 1 \frac{- 1}{1 + x}}$

Move the negative in front of the fraction.

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{3 - \frac{3 x}{1 + x} - 1 (- \frac{1}{1 + x})}$

Multiply $- 1 (- \frac{1}{1 + x})$ .

Multiply $- 1$ by $- 1$ .

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{3 - \frac{3 x}{1 + x} + 1 (\frac{1}{1 + x})}$

Multiply $\frac{1}{1 + x}$ by $1$ .

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{3 - \frac{3 x}{1 + x} + \frac{1}{1 + x}}$

Combine the numerators over the common denominator.

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{3 + \frac{- 3 x + 1}{1 + x}}$

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

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{\frac{3 (1 + x)}{1 + x} + \frac{- 3 x + 1}{1 + x}}$

Combine the numerators over the common denominator.

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{\frac{3 (1 + x) - 3 x + 1}{1 + x}}$

Reorder terms.

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{\frac{- 3 x + 3 (1 + x) + 1}{x + 1}}$

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

Apply the distributive property.

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{\frac{- 3 x + 3 \cdot 1 + 3 x + 1}{x + 1}}$

Multiply $3$ by $1$ .

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{\frac{- 3 x + 3 + 3 x + 1}{x + 1}}$

Add $- 3 x$ and $3 x$ .

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{\frac{0 + 3 + 1}{x + 1}}$

Add $0$ and $3$ .

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{\frac{3 + 1}{x + 1}}$

Add $3$ and $1$ .

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{\frac{4 x}{x + 1}}{\frac{4}{x + 1}}$

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{4 x}{x + 1} \cdot \frac{x + 1}{4}$

Cancel the common factor of $4$ .

Factor $4$ out of $4 x$ .

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{4 (x)}{x + 1} \cdot \frac{x + 1}{4}$

Cancel the common factor.

$f (\frac{3 x}{2 (1 + x)} - \frac{1}{2 (1 + x)}) = \frac{4 x}{x + 1} \cdot \frac{x + 1}{4}$

Rewrite the expression.

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

Cancel the common factor of $x + 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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