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

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

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.

$y + 2, 1$

Remove parentheses.

$y + 2, 1$

The LCM of one and any expression is the expression.

$y + 2$

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

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

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

Simplify the left side.

Cancel the common factor of $y + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Apply the distributive property.

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

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

$3 - y = x y + 2 x$

Solve the equation.

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

$3 - y - x y = 2 x$

Subtract $3$ from both sides of the equation.

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

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

Factor $y$ out of $- y$ .

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

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

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

Factor $y$ out of $y \cdot - 1 + y (- x)$ .

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Combine the numerators over the common denominator.

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

Rewrite $- 1$ as $- 1 (1)$ .

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

Factor $- 1$ out of $- x$ .

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

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

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

Move the negative in front of the fraction.

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

Step 3

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

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

Step 4

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

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

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

Remove parentheses.

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

Multiply the numerator and denominator of the complex fraction by $x + 2$ .

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

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

Combine.

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

Apply the distributive property.

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

Cancel the common factor of $x + 2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

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

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

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

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

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

Apply the distributive property.

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

Multiply $2$ by $3$ .

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

Multiply $- 1$ by $2$ .

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

Apply the distributive property.

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

Multiply $- 3$ by $2$ .

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

Subtract $6$ from $6$ .

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

Add $- 2 x - 3 x$ and $0$ .

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

Subtract $3 x$ from $- 2 x$ .

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

Move the negative in front of the fraction.

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

Remove unnecessary parentheses.

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

Factor out negative.

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

Simplify the denominator.

Multiply $x + 2$ by $1$ .

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

Subtract $x$ from $x$ .

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

Add $0$ and $2$ .

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

Add $2$ and $3$ .

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

Reduce the expression by cancelling the common factors.

Factor $\frac{5 x}{x + 2}$ out of $\frac{- (x + 2) \frac{5 x}{x + 2}}{5}$ .

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

Cancel the common factor of $5$ .

Factor $5$ out of $5 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x + 2$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the expression.

Move $- 1$ to the left of $x$ .

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

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

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

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

Set up the composite result function.

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

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

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

Multiply the numerator and denominator of the complex fraction by $1 + x$ .

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

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

Combine.

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

Apply the distributive property.

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

Cancel the common factor of $1 + x$ .

Move the leading negative in $- \frac{2 x - 3}{1 + x}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

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

Factor $1 + x$ out of $(1 + x) \cdot 3$ .

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

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

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

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

Multiply $- 1$ by $- 1$ .

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

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

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

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

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

Apply the distributive property.

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

Multiply $3$ by $1$ .

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

Add $2 x$ and $3 x$ .

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

Subtract $3$ from $3$ .

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

Add $5 x$ and $0$ .

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

Simplify the denominator.

Apply the distributive property.

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

Multiply $2$ by $- 1$ .

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

Multiply $- 1$ by $- 3$ .

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

Apply the distributive property.

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

Multiply $2$ by $1$ .

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

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

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

Add $- 2 x$ and $2 x$ .

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

Add $0$ and $3$ .

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

Add $3$ and $2$ .

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

Simplify terms.

Factor $\frac{5 x}{x + 1}$ out of $\frac{(1 + x) \frac{5 x}{x + 1}}{5}$ .

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

Cancel the common factor of $5$ .

Factor $5$ out of $5 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Cancel the common factor of $1 + x$ and $x + 1$ .

Reorder terms.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

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

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