Find the Inverse f(x)=(3x+2)/(2x+3)

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

Step 1

Write $f (x) = \frac{3 x + 2}{2 x + 3}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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

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

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.

$2 y + 3, 1$

Remove parentheses.

$2 y + 3, 1$

The LCM of one and any expression is the expression.

$2 y + 3$

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Apply the distributive property.

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

Reorder.

Rewrite using the commutative property of multiplication.

$3 y + 2 = 2 x y + x \cdot 3$

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

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

Solve the equation.

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

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

Subtract $2$ from both sides of the equation.

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

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

Factor $y$ out of $3 y$ .

$y \cdot 3 - 2 x y = 3 x - 2$

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

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

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

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Combine the numerators over the common denominator.

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

Step 4

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

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

Step 5

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

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

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

Simplify the numerator.

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

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

Apply the distributive property.

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

Multiply $3$ by $3$ .

$f^{- 1} (\frac{3 x + 2}{2 x + 3}) = \frac{\frac{9 x + 3 \cdot 2 - 2 (2 x + 3)}{2 x + 3}}{3 - 2 \frac{3 x + 2}{2 x + 3}}$

Multiply $3$ by $2$ .

$f^{- 1} (\frac{3 x + 2}{2 x + 3}) = \frac{\frac{9 x + 6 - 2 (2 x + 3)}{2 x + 3}}{3 - 2 \frac{3 x + 2}{2 x + 3}}$

Apply the distributive property.

$f^{- 1} (\frac{3 x + 2}{2 x + 3}) = \frac{\frac{9 x + 6 - 2 (2 x) - 2 \cdot 3}{2 x + 3}}{3 - 2 \frac{3 x + 2}{2 x + 3}}$

Multiply $2$ by $- 2$ .

$f^{- 1} (\frac{3 x + 2}{2 x + 3}) = \frac{\frac{9 x + 6 - 4 x - 2 \cdot 3}{2 x + 3}}{3 - 2 \frac{3 x + 2}{2 x + 3}}$

Multiply $- 2$ by $3$ .

$f^{- 1} (\frac{3 x + 2}{2 x + 3}) = \frac{\frac{9 x + 6 - 4 x - 6}{2 x + 3}}{3 - 2 \frac{3 x + 2}{2 x + 3}}$

Subtract $4 x$ from $9 x$ .

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

Subtract $6$ from $6$ .

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

Add $5 x$ and $0$ .

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

Simplify the denominator.

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

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

Move the negative in front of the fraction.

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

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

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

Combine the numerators over the common denominator.

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

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

Apply the distributive property.

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

Multiply $2$ by $3$ .

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

Multiply $3$ by $3$ .

$f^{- 1} (\frac{3 x + 2}{2 x + 3}) = \frac{\frac{5 x}{2 x + 3}}{\frac{6 x + 9 - 2 (3 x + 2)}{2 x + 3}}$

Apply the distributive property.

$f^{- 1} (\frac{3 x + 2}{2 x + 3}) = \frac{\frac{5 x}{2 x + 3}}{\frac{6 x + 9 - 2 (3 x) - 2 \cdot 2}{2 x + 3}}$

Multiply $3$ by $- 2$ .

$f^{- 1} (\frac{3 x + 2}{2 x + 3}) = \frac{\frac{5 x}{2 x + 3}}{\frac{6 x + 9 - 6 x - 2 \cdot 2}{2 x + 3}}$

Multiply $- 2$ by $2$ .

$f^{- 1} (\frac{3 x + 2}{2 x + 3}) = \frac{\frac{5 x}{2 x + 3}}{\frac{6 x + 9 - 6 x - 4}{2 x + 3}}$

Subtract $6 x$ from $6 x$ .

$f^{- 1} (\frac{3 x + 2}{2 x + 3}) = \frac{\frac{5 x}{2 x + 3}}{\frac{0 + 9 - 4}{2 x + 3}}$

Add $0$ and $9$ .

$f^{- 1} (\frac{3 x + 2}{2 x + 3}) = \frac{\frac{5 x}{2 x + 3}}{\frac{9 - 4}{2 x + 3}}$

Subtract $4$ from $9$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $5$ .

Factor $5$ out of $5 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

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

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

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

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

Apply the distributive property.

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

Multiply $3$ by $3$ .

$f (\frac{3 x - 2}{3 - 2 x}) = \frac{\frac{9 x + 3 \cdot - 2 + 2 (3 - 2 x)}{- 2 x + 3}}{2 (\frac{3 x - 2}{3 - 2 x}) + 3}$

Multiply $3$ by $- 2$ .

$f (\frac{3 x - 2}{3 - 2 x}) = \frac{\frac{9 x - 6 + 2 (3 - 2 x)}{- 2 x + 3}}{2 (\frac{3 x - 2}{3 - 2 x}) + 3}$

Apply the distributive property.

$f (\frac{3 x - 2}{3 - 2 x}) = \frac{\frac{9 x - 6 + 2 \cdot 3 + 2 (- 2 x)}{- 2 x + 3}}{2 (\frac{3 x - 2}{3 - 2 x}) + 3}$

Multiply $2$ by $3$ .

$f (\frac{3 x - 2}{3 - 2 x}) = \frac{\frac{9 x - 6 + 6 + 2 (- 2 x)}{- 2 x + 3}}{2 (\frac{3 x - 2}{3 - 2 x}) + 3}$

Multiply $- 2$ by $2$ .

$f (\frac{3 x - 2}{3 - 2 x}) = \frac{\frac{9 x - 6 + 6 - 4 x}{- 2 x + 3}}{2 (\frac{3 x - 2}{3 - 2 x}) + 3}$

Subtract $4 x$ from $9 x$ .

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

Add $- 6$ and $6$ .

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

Add $5 x$ and $0$ .

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

Simplify the denominator.

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

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

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

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

Combine the numerators over the common denominator.

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

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

Apply the distributive property.

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

Multiply $3$ by $2$ .

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

Multiply $2$ by $- 2$ .

$f (\frac{3 x - 2}{3 - 2 x}) = \frac{\frac{5 x}{- 2 x + 3}}{\frac{6 x - 4 + 3 (3 - 2 x)}{3 - 2 x}}$

Apply the distributive property.

$f (\frac{3 x - 2}{3 - 2 x}) = \frac{\frac{5 x}{- 2 x + 3}}{\frac{6 x - 4 + 3 \cdot 3 + 3 (- 2 x)}{3 - 2 x}}$

Multiply $3$ by $3$ .

$f (\frac{3 x - 2}{3 - 2 x}) = \frac{\frac{5 x}{- 2 x + 3}}{\frac{6 x - 4 + 9 + 3 (- 2 x)}{3 - 2 x}}$

Multiply $- 2$ by $3$ .

$f (\frac{3 x - 2}{3 - 2 x}) = \frac{\frac{5 x}{- 2 x + 3}}{\frac{6 x - 4 + 9 - 6 x}{3 - 2 x}}$

Subtract $6 x$ from $6 x$ .

$f (\frac{3 x - 2}{3 - 2 x}) = \frac{\frac{5 x}{- 2 x + 3}}{\frac{0 - 4 + 9}{3 - 2 x}}$

Subtract $4$ from $0$ .

$f (\frac{3 x - 2}{3 - 2 x}) = \frac{\frac{5 x}{- 2 x + 3}}{\frac{- 4 + 9}{3 - 2 x}}$

Add $- 4$ and $9$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $5$ .

Factor $5$ out of $5 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

Reorder terms.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse f(x)=(3x+2)/(2x+3)

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