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

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

Write $K (x) = \frac{2 x + 1}{3 x - 2}$ as an equation.

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

Interchange the variables.

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

Solve for $y$ .

Multiply the equation by $3 y - 2$ .

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

Simplify the left side.

Apply the distributive property.

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

Simplify the right side.

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

Cancel the common factor.

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

Rewrite the expression.

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

Solve for $y$ .

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Combine the numerators over the common denominator.

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

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

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

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

To verify the inverse, check if $K^{- 1} (K (x)) = x$ and $K (K^{- 1} (x)) = x$ .

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

Set up the composite result function.

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

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

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

Simplify the numerator.

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

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

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

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

Apply the distributive property.

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

Multiply $2$ by $2$ .

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

Multiply $2$ by $1$ .

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

Add $3 x$ and $4 x$ .

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x + 2 - 2}{3 x - 2}}{3 (\frac{2 x + 1}{3 x - 2}) - 2}$

Subtract $2$ from $2$ .

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x + 0}{3 x - 2}}{3 (\frac{2 x + 1}{3 x - 2}) - 2}$

Add $7 x$ and $0$ .

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{3 (\frac{2 x + 1}{3 x - 2}) - 2}$

Simplify the denominator.

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

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 (2 x + 1)}{3 x - 2} - 2}$

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

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 (2 x + 1)}{3 x - 2} - 2 \cdot \frac{3 x - 2}{3 x - 2}}$

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

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 (2 x + 1)}{3 x - 2} + \frac{- 2 (3 x - 2)}{3 x - 2}}$

Combine the numerators over the common denominator.

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 (2 x + 1) - 2 (3 x - 2)}{3 x - 2}}$

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

Apply the distributive property.

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 (2 x) + 3 \cdot 1 - 2 (3 x - 2)}{3 x - 2}}$

Multiply $2$ by $3$ .

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{6 x + 3 \cdot 1 - 2 (3 x - 2)}{3 x - 2}}$

Multiply $3$ by $1$ .

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{6 x + 3 - 2 (3 x - 2)}{3 x - 2}}$

Apply the distributive property.

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{6 x + 3 - 2 (3 x) - 2 \cdot - 2}{3 x - 2}}$

Multiply $3$ by $- 2$ .

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{6 x + 3 - 6 x - 2 \cdot - 2}{3 x - 2}}$

Multiply $- 2$ by $- 2$ .

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{6 x + 3 - 6 x + 4}{3 x - 2}}$

Subtract $6 x$ from $6 x$ .

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{0 + 3 + 4}{3 x - 2}}$

Add $0$ and $3$ .

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 + 4}{3 x - 2}}$

Add $3$ and $4$ .

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{7}{3 x - 2}}$

Multiply the numerator by the reciprocal of the denominator.

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{7 x}{3 x - 2} \cdot \frac{3 x - 2}{7}$

Cancel the common factor of $7$ .

Factor $7$ out of $7 x$ .

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{7 (x)}{3 x - 2} \cdot \frac{3 x - 2}{7}$

Cancel the common factor.

$K^{- 1} (\frac{2 x + 1}{3 x - 2}) = \frac{7 x}{3 x - 2} \cdot \frac{3 x - 2}{7}$

Rewrite the expression.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

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

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

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

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

Apply the distributive property.

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

Multiply $2$ by $1$ .

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

Multiply $2$ by $2$ .

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

Add $3 x$ and $4 x$ .

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x + 2 - 2}{3 x - 2}}{3 (\frac{1 + 2 x}{3 x - 2}) - 2}$

Subtract $2$ from $2$ .

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x + 0}{3 x - 2}}{3 (\frac{1 + 2 x}{3 x - 2}) - 2}$

Add $7 x$ and $0$ .

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{3 (\frac{1 + 2 x}{3 x - 2}) - 2}$

Simplify the denominator.

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

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 (1 + 2 x)}{3 x - 2} - 2}$

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

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 (1 + 2 x)}{3 x - 2} - 2 \cdot \frac{3 x - 2}{3 x - 2}}$

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

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 (1 + 2 x)}{3 x - 2} + \frac{- 2 (3 x - 2)}{3 x - 2}}$

Combine the numerators over the common denominator.

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 (1 + 2 x) - 2 (3 x - 2)}{3 x - 2}}$

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

Apply the distributive property.

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 \cdot 1 + 3 (2 x) - 2 (3 x - 2)}{3 x - 2}}$

Multiply $3$ by $1$ .

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 + 3 (2 x) - 2 (3 x - 2)}{3 x - 2}}$

Multiply $2$ by $3$ .

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 + 6 x - 2 (3 x - 2)}{3 x - 2}}$

Apply the distributive property.

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 + 6 x - 2 (3 x) - 2 \cdot - 2}{3 x - 2}}$

Multiply $3$ by $- 2$ .

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 + 6 x - 6 x - 2 \cdot - 2}{3 x - 2}}$

Multiply $- 2$ by $- 2$ .

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{3 + 6 x - 6 x + 4}{3 x - 2}}$

Add $3$ and $4$ .

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{6 x - 6 x + 7}{3 x - 2}}$

Subtract $6 x$ from $6 x$ .

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{0 + 7}{3 x - 2}}$

Add $0$ and $7$ .

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{\frac{7 x}{3 x - 2}}{\frac{7}{3 x - 2}}$

Multiply the numerator by the reciprocal of the denominator.

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{7 x}{3 x - 2} \cdot \frac{3 x - 2}{7}$

Cancel the common factor of $7$ .

Factor $7$ out of $7 x$ .

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{7 (x)}{3 x - 2} \cdot \frac{3 x - 2}{7}$

Cancel the common factor.

$K (\frac{1 + 2 x}{3 x - 2}) = \frac{7 x}{3 x - 2} \cdot \frac{3 x - 2}{7}$

Rewrite the expression.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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