Find the Inverse y=(2+3x)/(7-x)

$y = \frac{2 + 3 x}{7 - x}$

Step 1

Interchange the variables.

$x = \frac{2 + 3 y}{7 - y}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{2 + 3 y}{7 - y} = x$ .

$\frac{2 + 3 y}{7 - y} = x$

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.

$7 - y, 1$

Remove parentheses.

$7 - y, 1$

The LCM of one and any expression is the expression.

$7 - y$

Multiply each term in $\frac{2 + 3 y}{7 - y} = x$ by $7 - y$ to eliminate the fractions.

Multiply each term in $\frac{2 + 3 y}{7 - y} = x$ by $7 - y$ .

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

Simplify the left side.

Cancel the common factor of $7 - y$ .

Cancel the common factor.

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

Rewrite the expression.

$2 + 3 y = x (7 - y)$

Simplify the right side.

Apply the distributive property.

$2 + 3 y = x \cdot 7 + x (- y)$

Reorder.

Move $7$ to the left of $x$ .

$2 + 3 y = 7 \cdot x + x (- y)$

Rewrite using the commutative property of multiplication.

$2 + 3 y = 7 x - x y$

Solve the equation.

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

$2 + 3 y + x y = 7 x$

Subtract $2$ from both sides of the equation.

$3 y + x y = 7 x - 2$

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

Factor $y$ out of $3 y$ .

$y \cdot 3 + x y = 7 x - 2$

Factor $y$ out of $x y$ .

$y \cdot 3 + y x = 7 x - 2$

Factor $y$ out of $y \cdot 3 + y x$ .

$y (3 + x) = 7 x - 2$

Divide each term in $y (3 + x) = 7 x - 2$ by $3 + x$ and simplify.

Divide each term in $y (3 + x) = 7 x - 2$ by $3 + x$ .

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

Simplify the left side.

Cancel the common factor of $3 + x$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

$y = \frac{7 x}{3 + x} + \frac{- 2}{3 + x}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{7 x - 2}{3 + x}$

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = \frac{7 x - 2}{3 + x}$ is the inverse of $f (x) = \frac{2 + 3 x}{7 - x}$ .

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 $f^{- 1} (\frac{2 + 3 x}{7 - x})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Remove parentheses.

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

Multiply the numerator and denominator of the complex fraction by $7 - x$ .

Multiply $\frac{7 \frac{2 + 3 x}{7 - x} - 2}{3 + \frac{2 + 3 x}{7 - x}}$ by $\frac{7 - x}{7 - x}$ .

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

Combine.

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

Apply the distributive property.

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

Cancel the common factor of $7 - x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Factor $7 - x$ out of $(7 - x) \cdot 7 \frac{2 + 3 x}{7 - x} + (7 - x) \cdot - 2$ .

Factor $7 - x$ out of $(7 - x) \cdot 7 \frac{2 + 3 x}{7 - x}$ .

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

Factor $7 - x$ out of $(7 - x) \cdot - 2$ .

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

Factor $7 - x$ out of $(7 - x) (7 \frac{2 + 3 x}{7 - x}) + (7 - x) (- 2)$ .

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

Combine $7$ and $\frac{2 + 3 x}{7 - x}$ .

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

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{7 - x}{7 - x}$ .

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

Combine $- 2$ and $\frac{7 - x}{7 - x}$ .

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

Rewrite $\frac{7 (2 + 3 x) - 2 (7 - x)}{- x + 7}$ in a factored form.

Apply the distributive property.

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

Multiply $7$ by $2$ .

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{(7 - x) (\frac{14 + 7 (3 x) - 2 (7 - x)}{- x + 7})}{(7 - x) \cdot 3 + 2 + 3 x}$

Multiply $3$ by $7$ .

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{(7 - x) (\frac{14 + 21 x - 2 (7 - x)}{- x + 7})}{(7 - x) \cdot 3 + 2 + 3 x}$

Apply the distributive property.

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{(7 - x) (\frac{14 + 21 x - 2 \cdot 7 - 2 (- x)}{- x + 7})}{(7 - x) \cdot 3 + 2 + 3 x}$

Multiply $- 2$ by $7$ .

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{(7 - x) (\frac{14 + 21 x - 14 - 2 (- x)}{- x + 7})}{(7 - x) \cdot 3 + 2 + 3 x}$

Multiply $- 1$ by $- 2$ .

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{(7 - x) (\frac{14 + 21 x - 14 + 2 x}{- x + 7})}{(7 - x) \cdot 3 + 2 + 3 x}$

Subtract $14$ from $14$ .

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{(7 - x) (\frac{21 x + 0 + 2 x}{- x + 7})}{(7 - x) \cdot 3 + 2 + 3 x}$

Add $21 x$ and $0$ .

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{(7 - x) (\frac{21 x + 2 x}{- x + 7})}{(7 - x) \cdot 3 + 2 + 3 x}$

Add $21 x$ and $2 x$ .

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{(7 - x) (\frac{23 x}{- x + 7})}{(7 - x) \cdot 3 + 2 + 3 x}$

Simplify the denominator.

Apply the distributive property.

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{(7 - x) (\frac{23 x}{- x + 7})}{7 \cdot 3 - x \cdot 3 + 2 + 3 x}$

Multiply $7$ by $3$ .

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{(7 - x) (\frac{23 x}{- x + 7})}{21 - x \cdot 3 + 2 + 3 x}$

Multiply $3$ by $- 1$ .

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{(7 - x) (\frac{23 x}{- x + 7})}{21 - 3 x + 2 + 3 x}$

Add $21$ and $2$ .

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{(7 - x) (\frac{23 x}{- x + 7})}{- 3 x + 23 + 3 x}$

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

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{(7 - x) (\frac{23 x}{- x + 7})}{0 + 23}$

Add $0$ and $23$ .

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{(7 - x) (\frac{23 x}{- x + 7})}{23}$

Simplify terms.

Factor $\frac{23 x}{- x + 7}$ out of $\frac{(7 - x) \frac{23 x}{- x + 7}}{23}$ .

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{23 x}{- x + 7} \cdot \frac{7 - x}{23}$

Cancel the common factor of $23$ .

Factor $23$ out of $23 x$ .

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{23 (x)}{- x + 7} \cdot \frac{7 - x}{23}$

Cancel the common factor.

$f^{- 1} (\frac{2 + 3 x}{7 - x}) = \frac{23 x}{- x + 7} \cdot \frac{7 - x}{23}$

Rewrite the expression.

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

Multiply $\frac{x}{- x + 7}$ by $7 - x$ .

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

Cancel the common factor of $7 - x$ and $- x + 7$ .

Reorder terms.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

Evaluate $f (\frac{7 x - 2}{3 + x})$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Multiply the numerator and denominator of the complex fraction by $3 + x$ .

Multiply $\frac{2 + 3 \frac{7 x - 2}{3 + x}}{7 - \frac{7 x - 2}{3 + x}}$ by $\frac{3 + x}{3 + x}$ .

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

Combine.

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

Apply the distributive property.

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

Cancel the common factor of $3 + x$ .

Move the leading negative in $- \frac{7 x - 2}{3 + x}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Factor $3 + x$ out of $(3 + x) \cdot 2 + (3 + x) \cdot 3 \frac{7 x - 2}{3 + x}$ .

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

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

Factor $3 + x$ out of $(3 + x) (2) + (3 + x) (3 \frac{7 x - 2}{3 + x})$ .

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

Combine $3$ and $\frac{7 x - 2}{3 + x}$ .

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

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

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

Combine the numerators over the common denominator.

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

Rewrite $\frac{2 (3 + x) + 3 (7 x - 2)}{3 + x}$ in a factored form.

Apply the distributive property.

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

Multiply $2$ by $3$ .

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

Apply the distributive property.

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

Multiply $7$ by $3$ .

$f (\frac{7 x - 2}{3 + x}) = \frac{(3 + x) (\frac{6 + 2 x + 21 x + 3 \cdot - 2}{3 + x})}{(3 + x) \cdot 7 - (7 x - 2)}$

Multiply $3$ by $- 2$ .

$f (\frac{7 x - 2}{3 + x}) = \frac{(3 + x) (\frac{6 + 2 x + 21 x - 6}{3 + x})}{(3 + x) \cdot 7 - (7 x - 2)}$

Subtract $6$ from $6$ .

$f (\frac{7 x - 2}{3 + x}) = \frac{(3 + x) (\frac{2 x + 21 x + 0}{3 + x})}{(3 + x) \cdot 7 - (7 x - 2)}$

Add $2 x + 21 x$ and $0$ .

$f (\frac{7 x - 2}{3 + x}) = \frac{(3 + x) (\frac{2 x + 21 x}{3 + x})}{(3 + x) \cdot 7 - (7 x - 2)}$

Add $2 x$ and $21 x$ .

$f (\frac{7 x - 2}{3 + x}) = \frac{(3 + x) (\frac{23 x}{3 + x})}{(3 + x) \cdot 7 - (7 x - 2)}$

Simplify the denominator.

Apply the distributive property.

$f (\frac{7 x - 2}{3 + x}) = \frac{(3 + x) (\frac{23 x}{3 + x})}{3 \cdot 7 + x \cdot 7 - (7 x - 2)}$

Multiply $3$ by $7$ .

$f (\frac{7 x - 2}{3 + x}) = \frac{(3 + x) (\frac{23 x}{3 + x})}{21 + x \cdot 7 - (7 x - 2)}$

Move $7$ to the left of $x$ .

$f (\frac{7 x - 2}{3 + x}) = \frac{(3 + x) (\frac{23 x}{3 + x})}{21 + 7 \cdot x - (7 x - 2)}$

Apply the distributive property.

$f (\frac{7 x - 2}{3 + x}) = \frac{(3 + x) (\frac{23 x}{3 + x})}{21 + 7 x - (7 x) + 2}$

Multiply $7$ by $- 1$ .

$f (\frac{7 x - 2}{3 + x}) = \frac{(3 + x) (\frac{23 x}{3 + x})}{21 + 7 x - 7 x + 2}$

Multiply $- 1$ by $- 2$ .

$f (\frac{7 x - 2}{3 + x}) = \frac{(3 + x) (\frac{23 x}{3 + x})}{21 + 7 x - 7 x + 2}$

Add $21$ and $2$ .

$f (\frac{7 x - 2}{3 + x}) = \frac{(3 + x) (\frac{23 x}{3 + x})}{7 x - 7 x + 23}$

Subtract $7 x$ from $7 x$ .

$f (\frac{7 x - 2}{3 + x}) = \frac{(3 + x) (\frac{23 x}{3 + x})}{0 + 23}$

Add $0$ and $23$ .

$f (\frac{7 x - 2}{3 + x}) = \frac{(3 + x) (\frac{23 x}{3 + x})}{23}$

Reduce the expression by cancelling the common factors.

Factor $\frac{23 x}{3 + x}$ out of $\frac{(3 + x) \frac{23 x}{3 + x}}{23}$ .

$f (\frac{7 x - 2}{3 + x}) = \frac{23 x}{3 + x} \cdot \frac{3 + x}{23}$

Cancel the common factor of $23$ .

Factor $23$ out of $23 x$ .

$f (\frac{7 x - 2}{3 + x}) = \frac{23 (x)}{3 + x} \cdot \frac{3 + x}{23}$

Cancel the common factor.

$f (\frac{7 x - 2}{3 + x}) = \frac{23 x}{3 + x} \cdot \frac{3 + x}{23}$

Rewrite the expression.

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

Cancel the common factor of $3 + x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{7 x - 2}{3 + x}$ is the inverse of $f (x) = \frac{2 + 3 x}{7 - x}$ .

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

Find the Inverse y=(2+3x)/(7-x)

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