Find the Inverse f(x)=(1-2x)/(5x-1)

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

Step 1

Write $f (x) = \frac{1 - 2 x}{5 x - 1}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

Multiply the equation by $5 y - 1$ .

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

Simplify the left side.

Simplify $(5 y - 1) (x)$ .

Apply the distributive property.

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

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

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

Simplify the right side.

Cancel the common factor of $5 y - 1$ .

Cancel the common factor.

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

Rewrite the expression.

$5 y x - x = 1 - 2 y$

Solve for $y$ .

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

$5 y x - x + 2 y = 1$

Add $x$ to both sides of the equation.

$5 y x + 2 y = 1 + x$

Factor $y$ out of $5 y x + 2 y$ .

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

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

Factor $y$ out of $2 y$ .

$y (5 x) + y \cdot 2 = 1 + x$

Factor $y$ out of $y (5 x) + y \cdot 2$ .

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

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

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

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

Simplify the left side.

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

Cancel the common factor.

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

Divide $y$ by $1$ .

$y = \frac{1}{5 x + 2} + \frac{x}{5 x + 2}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{1 + x}{5 x + 2}$

Step 4

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

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

Step 5

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

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

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

Remove parentheses.

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

Multiply the numerator and denominator of the complex fraction by $5 x - 1$ .

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

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

Combine.

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

Apply the distributive property.

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

Cancel the common factor of $5 x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Multiply $5 x - 1$ by $1$ .

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

Subtract $2 x$ from $5 x$ .

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

Add $- 1$ and $1$ .

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

Add $3 x$ and $0$ .

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

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

Simplify the denominator.

Factor $5 x - 1$ out of $(5 x - 1) \cdot 5 \frac{1 - 2 x}{5 x - 1} + (5 x - 1) \cdot 2$ .

Factor $5 x - 1$ out of $(5 x - 1) \cdot 2$ .

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

Factor $5 x - 1$ out of $(5 x - 1) (5 \frac{1 - 2 x}{5 x - 1}) + (5 x - 1) (2)$ .

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

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

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

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

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

Combine the numerators over the common denominator.

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

Rewrite $\frac{5 (1 - 2 x) + 2 (5 x - 1)}{5 x - 1}$ in a factored form.

Apply the distributive property.

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

Multiply $5$ by $1$ .

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

Multiply $- 2$ by $5$ .

$f^{- 1} (\frac{1 - 2 x}{5 x - 1}) = \frac{3 x}{(5 x - 1) (\frac{5 - 10 x + 2 (5 x - 1)}{5 x - 1})}$

Apply the distributive property.

$f^{- 1} (\frac{1 - 2 x}{5 x - 1}) = \frac{3 x}{(5 x - 1) (\frac{5 - 10 x + 2 (5 x) + 2 \cdot - 1}{5 x - 1})}$

Multiply $5$ by $2$ .

$f^{- 1} (\frac{1 - 2 x}{5 x - 1}) = \frac{3 x}{(5 x - 1) (\frac{5 - 10 x + 10 x + 2 \cdot - 1}{5 x - 1})}$

Multiply $2$ by $- 1$ .

$f^{- 1} (\frac{1 - 2 x}{5 x - 1}) = \frac{3 x}{(5 x - 1) (\frac{5 - 10 x + 10 x - 2}{5 x - 1})}$

Subtract $2$ from $5$ .

$f^{- 1} (\frac{1 - 2 x}{5 x - 1}) = \frac{3 x}{(5 x - 1) (\frac{- 10 x + 10 x + 3}{5 x - 1})}$

Add $- 10 x$ and $10 x$ .

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

Add $0$ and $3$ .

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

Factor $\frac{3}{5 x - 1}$ out of $\frac{3 x}{(5 x - 1) \frac{3}{5 x - 1}}$ .

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

Cancel the common factor of $5 x - 1$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $3$ .

Factor $3$ out of $3 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

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

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

Move the negative in front of the fraction.

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

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

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

Combine the numerators over the common denominator.

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

Rewrite $\frac{5 x + 2 - 2 (1 + x)}{5 x + 2}$ in a factored form.

Apply the distributive property.

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

Multiply $- 2$ by $1$ .

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

Subtract $2 x$ from $5 x$ .

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

Subtract $2$ from $2$ .

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

Add $3 x$ and $0$ .

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

Simplify the denominator.

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Rewrite $\frac{5 (1 + x) - (5 x + 2)}{5 x + 2}$ in a factored form.

Apply the distributive property.

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

Multiply $5$ by $1$ .

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

Apply the distributive property.

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

Multiply $5$ by $- 1$ .

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

Multiply $- 1$ by $2$ .

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

Subtract $2$ from $5$ .

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

Subtract $5 x$ from $5 x$ .

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

Add $0$ and $3$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $3$ .

Factor $3$ out of $3 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Find the Inverse f(x)=(1-2x)/(5x-1)

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