Find the Inverse f(x)=(7-5x)/(5x+2)

$f (x) = \frac{7 - 5 x}{5 x + 2}$

Step 1

Write $f (x) = \frac{7 - 5 x}{5 x + 2}$ as an equation.

$y = \frac{7 - 5 x}{5 x + 2}$

Step 2

Interchange the variables.

$x = \frac{7 - 5 y}{5 y + 2}$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{7 - 5 y}{5 y + 2} = x$ .

$\frac{7 - 5 y}{5 y + 2} = x$

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.

$5 y + 2, 1$

Remove parentheses.

$5 y + 2, 1$

The LCM of one and any expression is the expression.

$5 y + 2$

Multiply each term in $\frac{7 - 5 y}{5 y + 2} = x$ by $5 y + 2$ to eliminate the fractions.

Multiply each term in $\frac{7 - 5 y}{5 y + 2} = x$ by $5 y + 2$ .

$\frac{7 - 5 y}{5 y + 2} (5 y + 2) = x (5 y + 2)$

Simplify the left side.

Cancel the common factor of $5 y + 2$ .

Cancel the common factor.

$\frac{7 - 5 y}{5 y + 2} (5 y + 2) = x (5 y + 2)$

Rewrite the expression.

$7 - 5 y = x (5 y + 2)$

Simplify the right side.

Apply the distributive property.

$7 - 5 y = x (5 y) + x \cdot 2$

Reorder.

Rewrite using the commutative property of multiplication.

$7 - 5 y = 5 x y + x \cdot 2$

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

$7 - 5 y = 5 x y + 2 x$

Solve the equation.

Subtract $5 x y$ from both sides of the equation.

$7 - 5 y - 5 x y = 2 x$

Subtract $7$ from both sides of the equation.

$- 5 y - 5 x y = 2 x - 7$

Factor $5 y$ out of $- 5 y - 5 x y$ .

Factor $5 y$ out of $- 5 y$ .

$5 y (- 1) - 5 x y = 2 x - 7$

Factor $5 y$ out of $- 5 x y$ .

$5 y (- 1) + 5 y (- x) = 2 x - 7$

Factor $5 y$ out of $5 y (- 1) + 5 y (- x)$ .

$5 y (- 1 - x) = 2 x - 7$

Divide each term in $5 y (- 1 - x) = 2 x - 7$ by $5 (- 1 - x)$ and simplify.

Divide each term in $5 y (- 1 - x) = 2 x - 7$ by $5 (- 1 - x)$ .

$\frac{5 y (- 1 - x)}{5 (- 1 - x)} = \frac{2 x}{5 (- 1 - x)} + \frac{- 7}{5 (- 1 - x)}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 y (- 1 - x)}{5 (- 1 - x)} = \frac{2 x}{5 (- 1 - x)} + \frac{- 7}{5 (- 1 - x)}$

Rewrite the expression.

$\frac{y (- 1 - x)}{- 1 - x} = \frac{2 x}{5 (- 1 - x)} + \frac{- 7}{5 (- 1 - x)}$

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

Cancel the common factor.

$\frac{y (- 1 - x)}{- 1 - x} = \frac{2 x}{5 (- 1 - x)} + \frac{- 7}{5 (- 1 - x)}$

Divide $y$ by $1$ .

$y = \frac{2 x}{5 (- 1 - x)} + \frac{- 7}{5 (- 1 - x)}$

Simplify the right side.

Move the negative in front of the fraction.

$y = \frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}$

Step 4

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

$f^{- 1} (x) = \frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}$

Step 5

Verify if $f^{- 1} (x) = \frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}$ is the inverse of $f (x) = \frac{7 - 5 x}{5 x + 2}$ .

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 $f^{- 1} (\frac{7 - 5 x}{5 x + 2})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{2 (\frac{7 - 5 x}{5 x + 2})}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))} - \frac{7}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{2 (\frac{7 - 5 x}{5 x + 2}) - 7}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Combine $2$ and $\frac{7 - 5 x}{5 x + 2}$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{\frac{2 (7 - 5 x)}{5 x + 2} - 7}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

To write $- 7$ as a fraction with a common denominator, multiply by $\frac{5 x + 2}{5 x + 2}$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{\frac{2 (7 - 5 x)}{5 x + 2} - 7 \cdot \frac{5 x + 2}{5 x + 2}}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Simplify terms.

Combine $- 7$ and $\frac{5 x + 2}{5 x + 2}$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{\frac{2 (7 - 5 x)}{5 x + 2} + \frac{- 7 (5 x + 2)}{5 x + 2}}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{\frac{2 (7 - 5 x) - 7 (5 x + 2)}{5 x + 2}}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Simplify the numerator.

Apply the distributive property.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{\frac{2 \cdot 7 + 2 (- 5 x) - 7 (5 x + 2)}{5 x + 2}}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Multiply $2$ by $7$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{\frac{14 + 2 (- 5 x) - 7 (5 x + 2)}{5 x + 2}}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Multiply $- 5$ by $2$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{\frac{14 - 10 x - 7 (5 x + 2)}{5 x + 2}}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Apply the distributive property.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{\frac{14 - 10 x - 7 (5 x) - 7 \cdot 2}{5 x + 2}}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Multiply $5$ by $- 7$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{\frac{14 - 10 x - 35 x - 7 \cdot 2}{5 x + 2}}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Multiply $- 7$ by $2$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{\frac{14 - 10 x - 35 x - 14}{5 x + 2}}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Subtract $14$ from $14$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{\frac{- 10 x - 35 x + 0}{5 x + 2}}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Add $- 10 x - 35 x$ and $0$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{\frac{- 10 x - 35 x}{5 x + 2}}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Subtract $35 x$ from $- 10 x$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{\frac{- 45 x}{5 x + 2}}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Move the negative in front of the fraction.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- \frac{45 x}{5 x + 2}}{5 (- 1 - (\frac{7 - 5 x}{5 x + 2}))}$

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = - \frac{45 x}{5 x + 2} \cdot \frac{1}{5 (- 1 - \frac{7 - 5 x}{5 x + 2})}$

Simplify terms.

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{45 x}{5 x + 2}$ into the numerator.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- 45 x}{5 x + 2} \cdot \frac{1}{5 (- 1 - \frac{7 - 5 x}{5 x + 2})}$

Factor $5$ out of $- 45 x$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{5 (- 9 x)}{5 x + 2} \cdot \frac{1}{5 (- 1 - \frac{7 - 5 x}{5 x + 2})}$

Cancel the common factor.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{5 (- 9 x)}{5 x + 2} \cdot \frac{1}{5 (- 1 - \frac{7 - 5 x}{5 x + 2})}$

Rewrite the expression.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- 9 x}{5 x + 2} \cdot \frac{1}{- 1 - \frac{7 - 5 x}{5 x + 2}}$

Multiply $\frac{- 9 x}{5 x + 2}$ by $\frac{1}{- 1 - \frac{7 - 5 x}{5 x + 2}}$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- 9 x}{(5 x + 2) (- 1 - \frac{7 - 5 x}{5 x + 2})}$

Simplify the denominator.

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{5 x + 2}{5 x + 2}$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- 9 x}{(5 x + 2) (- 1 \cdot \frac{5 x + 2}{5 x + 2} - \frac{7 - 5 x}{5 x + 2})}$

Combine $- 1$ and $\frac{5 x + 2}{5 x + 2}$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- 9 x}{(5 x + 2) (\frac{- (5 x + 2)}{5 x + 2} - \frac{7 - 5 x}{5 x + 2})}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- 9 x}{(5 x + 2) (\frac{- (5 x + 2) - (7 - 5 x)}{5 x + 2})}$

Rewrite $\frac{- (5 x + 2) - (7 - 5 x)}{5 x + 2}$ in a factored form.

Factor $- 1$ out of $- (5 x + 2) - (7 - 5 x)$ .

Reorder $7$ and $- 5 x$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- 9 x}{(5 x + 2) (\frac{- (5 x + 2) - (- 5 x + 7)}{5 x + 2})}$

Factor $- 1$ out of $- (5 x + 2) - (- 5 x + 7)$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- 9 x}{(5 x + 2) (\frac{- (5 x + 2 - 5 x + 7)}{5 x + 2})}$

Subtract $5 x$ from $5 x$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- 9 x}{(5 x + 2) (\frac{- (0 + 2 + 7)}{5 x + 2})}$

Add $0$ and $2$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- 9 x}{(5 x + 2) (\frac{- (2 + 7)}{5 x + 2})}$

Add $2$ and $7$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- 9 x}{(5 x + 2) (\frac{- 1 \cdot 9}{5 x + 2})}$

Multiply $- 1$ by $9$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- 9 x}{(5 x + 2) (\frac{- 9}{5 x + 2})}$

Move the negative in front of the fraction.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- 9 x}{(5 x + 2) \cdot (- 1 \frac{9}{5 x + 2})}$

Factor out negative.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- 9 x}{- (5 x + 2) \frac{9}{5 x + 2}}$

Factor $\frac{9}{5 x + 2}$ out of $\frac{- 9 x}{- (5 x + 2) \frac{9}{5 x + 2}}$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{5 x + 2}{9} \cdot \frac{- 9 x}{- (5 x + 2)}$

Cancel the common factor of $5 x + 2$ .

Factor $5 x + 2$ out of $- (5 x + 2)$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{5 x + 2}{9} \cdot \frac{- 9 x}{(5 x + 2) \cdot - 1}$

Cancel the common factor.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{5 x + 2}{9} \cdot \frac{- 9 x}{(5 x + 2) \cdot - 1}$

Rewrite the expression.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{1}{9} \cdot \frac{- 9 x}{- 1}$

Cancel the common factor of $9$ .

Factor $9$ out of $- 9 x$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{1}{9} \cdot \frac{9 (- x)}{- 1}$

Cancel the common factor.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{1}{9} \cdot \frac{9 (- x)}{- 1}$

Rewrite the expression.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{- x}{- 1}$

Dividing two negative values results in a positive value.

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = \frac{x}{1}$

Divide $x$ by $1$ .

$f^{- 1} (\frac{7 - 5 x}{5 x + 2}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{7 - 5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)})}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Simplify the numerator.

Apply the distributive property.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{7 - 5 \frac{2 x}{5 (- 1 - x)} - 5 (- \frac{7}{5 (- 1 - x)})}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Cancel the common factor of $5$ .

Factor $5$ out of $- 5$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{7 + 5 (- 1) (\frac{2 x}{5 (- 1 - x)}) - 5 (- \frac{7}{5 (- 1 - x)})}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Cancel the common factor.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{7 + 5 \cdot (- 1 \frac{2 x}{5 (- 1 - x)}) - 5 (- \frac{7}{5 (- 1 - x)})}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Rewrite the expression.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{7 - 1 \frac{2 x}{- 1 - x} - 5 (- \frac{7}{5 (- 1 - x)})}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{7}{5 (- 1 - x)}$ into the numerator.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{7 - 1 \frac{2 x}{- 1 - x} - 5 \frac{- 7}{5 (- 1 - x)}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Factor $5$ out of $- 5$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{7 - 1 \frac{2 x}{- 1 - x} + 5 (- 1) (\frac{- 7}{5 (- 1 - x)})}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Cancel the common factor.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{7 - 1 \frac{2 x}{- 1 - x} + 5 \cdot (- 1 \frac{- 7}{5 (- 1 - x)})}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Rewrite the expression.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{7 - 1 \frac{2 x}{- 1 - x} - 1 \frac{- 7}{- 1 - x}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Simplify each term.

Rewrite $- 1 \frac{2 x}{- 1 - x}$ as $- \frac{2 x}{- 1 - x}$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{7 - \frac{2 x}{- 1 - x} - 1 \frac{- 7}{- 1 - x}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Move the negative in front of the fraction.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{7 - \frac{2 x}{- 1 - x} - 1 (- \frac{7}{- 1 - x})}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Multiply $- 1 (- \frac{7}{- 1 - x})$ .

Multiply $- 1$ by $- 1$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{7 - \frac{2 x}{- 1 - x} + 1 (\frac{7}{- 1 - x})}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Multiply $\frac{7}{- 1 - x}$ by $1$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{7 - \frac{2 x}{- 1 - x} + \frac{7}{- 1 - x}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Combine the numerators over the common denominator.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{7 + \frac{- 2 x + 7}{- 1 - x}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

To write $7$ as a fraction with a common denominator, multiply by $\frac{- 1 - x}{- 1 - x}$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{\frac{7 (- 1 - x)}{- 1 - x} + \frac{- 2 x + 7}{- 1 - x}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Combine the numerators over the common denominator.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{\frac{7 (- 1 - x) - 2 x + 7}{- 1 - x}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Reorder terms.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{\frac{- 2 x + 7 (- 1 - x) + 7}{- x - 1}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Rewrite $\frac{- 2 x + 7 (- 1 - x) + 7}{- x - 1}$ in a factored form.

Apply the distributive property.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{\frac{- 2 x + 7 \cdot - 1 + 7 (- x) + 7}{- x - 1}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Multiply $7$ by $- 1$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{\frac{- 2 x - 7 + 7 (- x) + 7}{- x - 1}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Multiply $- 1$ by $7$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{\frac{- 2 x - 7 - 7 x + 7}{- x - 1}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Subtract $7 x$ from $- 2 x$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{\frac{- 9 x - 7 + 7}{- x - 1}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Add $- 7$ and $7$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{\frac{- 9 x + 0}{- x - 1}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Add $- 9 x$ and $0$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{\frac{- 9 x}{- x - 1}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Move the negative in front of the fraction.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{5 (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) + 2}$

Simplify the denominator.

Apply the distributive property.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{5 (\frac{2 x}{5 (- 1 - x)}) + 5 (- \frac{7}{5 (- 1 - x)}) + 2}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{5 (\frac{2 x}{5 (- 1 - x)}) + 5 (- \frac{7}{5 (- 1 - x)}) + 2}$

Rewrite the expression.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{\frac{2 x}{- 1 - x} + 5 (- \frac{7}{5 (- 1 - x)}) + 2}$

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{7}{5 (- 1 - x)}$ into the numerator.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{\frac{2 x}{- 1 - x} + 5 (\frac{- 7}{5 (- 1 - x)}) + 2}$

Cancel the common factor.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{\frac{2 x}{- 1 - x} + 5 (\frac{- 7}{5 (- 1 - x)}) + 2}$

Rewrite the expression.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{\frac{2 x}{- 1 - x} + \frac{- 7}{- 1 - x} + 2}$

Combine the numerators over the common denominator.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{\frac{2 x - 7}{- 1 - x} + 2}$

To write $2$ as a fraction with a common denominator, multiply by $\frac{- 1 - x}{- 1 - x}$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{\frac{2 x - 7}{- 1 - x} + \frac{2 (- 1 - x)}{- 1 - x}}$

Combine the numerators over the common denominator.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{\frac{2 x - 7 + 2 (- 1 - x)}{- 1 - x}}$

Reorder terms.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{\frac{2 x + 2 (- 1 - x) - 7}{- x - 1}}$

Rewrite $\frac{2 x + 2 (- 1 - x) - 7}{- x - 1}$ in a factored form.

Apply the distributive property.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{\frac{2 x + 2 \cdot - 1 + 2 (- x) - 7}{- x - 1}}$

Multiply $2$ by $- 1$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{\frac{2 x - 2 + 2 (- x) - 7}{- x - 1}}$

Multiply $- 1$ by $2$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{\frac{2 x - 2 - 2 x - 7}{- x - 1}}$

Subtract $2 x$ from $2 x$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{\frac{0 - 2 - 7}{- x - 1}}$

Subtract $2$ from $0$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{\frac{- 2 - 7}{- x - 1}}$

Subtract $7$ from $- 2$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{\frac{- 9}{- x - 1}}$

Move the negative in front of the fraction.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{- \frac{9 x}{- x - 1}}{- \frac{9}{- x - 1}}$

Dividing two negative values results in a positive value.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{\frac{9 x}{- x - 1}}{\frac{9}{- x - 1}}$

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{9 x}{- x - 1} \cdot \frac{- x - 1}{9}$

Cancel the common factor of $9$ .

Factor $9$ out of $9 x$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{9 (x)}{- x - 1} \cdot \frac{- x - 1}{9}$

Cancel the common factor.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{9 x}{- x - 1} \cdot \frac{- x - 1}{9}$

Rewrite the expression.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{x}{- x - 1} \cdot (- x - 1)$

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

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{x (- x - 1)}{- x - 1}$

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

Cancel the common factor.

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = \frac{x (- x - 1)}{- x - 1}$

Divide $x$ by $1$ .

$f (\frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}$ is the inverse of $f (x) = \frac{7 - 5 x}{5 x + 2}$ .

$f^{- 1} (x) = \frac{2 x}{5 (- 1 - x)} - \frac{7}{5 (- 1 - x)}$

Find the Inverse f(x)=(7-5x)/(5x+2)

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