Find the Inverse -2/3x+2

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Simplify each term.

Combine $y$ and $\frac{2}{3}$ .

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

Move $2$ to the left of $y$ .

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

Subtract $2$ from both sides of the equation.

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

Multiply both sides of the equation by $- \frac{3}{2}$ .

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

Simplify both sides of the equation.

Simplify the left side.

Simplify $- \frac{3}{2} (- \frac{2 y}{3})$ .

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{3}{2}$ into the numerator.

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

Move the leading negative in $- \frac{2 y}{3}$ into the numerator.

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

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply.

Multiply $- 1$ by $- 1$ .

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

Multiply $y$ by $1$ .

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

Simplify the right side.

Simplify $- \frac{3}{2} (x - 2)$ .

Simplify terms.

Apply the distributive property.

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

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

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3}{2}$ into the numerator.

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

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 3$ by $- 1$ .

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

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

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

Step 3

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

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

Step 4

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

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

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

Simplify each term.

Simplify the numerator.

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

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

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

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

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

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

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

Factor $2$ out of $2$ .

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

Factor $2$ out of $2 (- \frac{x}{3}) + 2 (1)$ .

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

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

Multiply $3$ by $2$ .

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

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

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

Combine the numerators over the common denominator.

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

Combine $6$ and $\frac{- x + 3}{3}$ .

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

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{6 (- x + 3)}{3}$ by cancelling the common factors.

Factor $3$ out of $6 (- x + 3)$ .

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

Factor $3$ out of $3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $2 (- x + 3)$ by $1$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $- x + 3$ by $1$ .

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

Apply the distributive property.

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

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 1$ by $3$ .

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

Combine the opposite terms in $x - 3 + 3$ .

Add $- 3$ and $3$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{2}{3}$ into the numerator.

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

Move the leading negative in $- \frac{3 x}{2}$ into the numerator.

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

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $3$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{2}{3}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Combine the opposite terms in $x - 2 + 2$ .

Add $- 2$ and $2$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse -2/3x+2

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