Find the Inverse (3+4x)/(1-5x)

$\frac{3 + 4 x}{1 - 5 x}$

Interchange the variables.

$x = \frac{3 + 4 y}{1 - 5 y}$

Solve for $y$ .

Rewrite the equation as $\frac{3 + 4 y}{1 - 5 y} = x$ .

$\frac{3 + 4 y}{1 - 5 y} = x$

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.

$1 - 5 y, 1$

Remove parentheses.

$1 - 5 y, 1$

The LCM of one and any expression is the expression.

$1 - 5 y$

Multiply each term in $\frac{3 + 4 y}{1 - 5 y} = x$ by $1 - 5 y$ to eliminate the fractions.

Multiply each term in $\frac{3 + 4 y}{1 - 5 y} = x$ by $1 - 5 y$ .

$\frac{3 + 4 y}{1 - 5 y} (1 - 5 y) = x (1 - 5 y)$

Simplify the left side.

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

Cancel the common factor.

$\frac{3 + 4 y}{1 - 5 y} (1 - 5 y) = x (1 - 5 y)$

Rewrite the expression.

$3 + 4 y = x (1 - 5 y)$

Simplify the right side.

Apply the distributive property.

$3 + 4 y = x \cdot 1 + x (- 5 y)$

Simplify the expression.

Multiply $x$ by $1$ .

$3 + 4 y = x + x (- 5 y)$

Rewrite using the commutative property of multiplication.

$3 + 4 y = x - 5 x y$

Solve the equation.

Add $5 x y$ to both sides of the equation.

$3 + 4 y + 5 x y = x$

Subtract $3$ from both sides of the equation.

$4 y + 5 x y = x - 3$

Factor $y$ out of $4 y + 5 x y$ .

Factor $y$ out of $4 y$ .

$y \cdot 4 + 5 x y = x - 3$

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

$y \cdot 4 + y (5 x) = x - 3$

Factor $y$ out of $y \cdot 4 + y (5 x)$ .

$y (4 + 5 x) = x - 3$

Divide each term in $y (4 + 5 x) = x - 3$ by $4 + 5 x$ and simplify.

Divide each term in $y (4 + 5 x) = x - 3$ by $4 + 5 x$ .

$\frac{y (4 + 5 x)}{4 + 5 x} = \frac{x}{4 + 5 x} + \frac{- 3}{4 + 5 x}$

Simplify the left side.

Cancel the common factor of $4 + 5 x$ .

Cancel the common factor.

$\frac{y (4 + 5 x)}{4 + 5 x} = \frac{x}{4 + 5 x} + \frac{- 3}{4 + 5 x}$

Divide $y$ by $1$ .

$y = \frac{x}{4 + 5 x} + \frac{- 3}{4 + 5 x}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{x - 3}{4 + 5 x}$

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

$f^{- 1} (x) = \frac{x - 3}{4 + 5 x}$

Verify if $f^{- 1} (x) = \frac{x - 3}{4 + 5 x}$ is the inverse of $f (x) = \frac{3 + 4 x}{1 - 5 x}$ .

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 $f^{- 1} (\frac{3 + 4 x}{1 - 5 x})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{(\frac{3 + 4 x}{1 - 5 x}) - 3}{4 + 5 (\frac{3 + 4 x}{1 - 5 x})}$

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

Multiply $\frac{\frac{3 + 4 x}{1 - 5 x} - 3}{4 + 5 \frac{3 + 4 x}{1 - 5 x}}$ by $\frac{1 - 5 x}{1 - 5 x}$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{1 - 5 x}{1 - 5 x} \cdot \frac{\frac{3 + 4 x}{1 - 5 x} - 3}{4 + 5 (\frac{3 + 4 x}{1 - 5 x})}$

Combine.

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{(1 - 5 x) (\frac{3 + 4 x}{1 - 5 x} - 3)}{(1 - 5 x) (4 + 5 (\frac{3 + 4 x}{1 - 5 x}))}$

Apply the distributive property.

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{(1 - 5 x) (\frac{3 + 4 x}{1 - 5 x}) + (1 - 5 x) \cdot - 3}{(1 - 5 x) \cdot 4 + (1 - 5 x) (5 (\frac{3 + 4 x}{1 - 5 x}))}$

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

Cancel the common factor.

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{(1 - 5 x) (\frac{3 + 4 x}{1 - 5 x}) + (1 - 5 x) \cdot - 3}{(1 - 5 x) \cdot 4 + (1 - 5 x) (5 (\frac{3 + 4 x}{1 - 5 x}))}$

Rewrite the expression.

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{3 + 4 x + (1 - 5 x) \cdot - 3}{(1 - 5 x) \cdot 4 + (1 - 5 x) (5 (\frac{3 + 4 x}{1 - 5 x}))}$

Simplify the numerator.

Apply the distributive property.

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{3 + 4 x + 1 \cdot - 3 - 5 x \cdot - 3}{(1 - 5 x) \cdot 4 + (1 - 5 x) \cdot (5 (\frac{3 + 4 x}{1 - 5 x}))}$

Multiply $- 3$ by $1$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{3 + 4 x - 3 - 5 x \cdot - 3}{(1 - 5 x) \cdot 4 + (1 - 5 x) \cdot (5 (\frac{3 + 4 x}{1 - 5 x}))}$

Multiply $- 3$ by $- 5$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{3 + 4 x - 3 + 15 x}{(1 - 5 x) \cdot 4 + (1 - 5 x) \cdot (5 (\frac{3 + 4 x}{1 - 5 x}))}$

Subtract $3$ from $3$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{4 x + 0 + 15 x}{(1 - 5 x) \cdot 4 + (1 - 5 x) \cdot (5 (\frac{3 + 4 x}{1 - 5 x}))}$

Add $4 x$ and $0$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{4 x + 15 x}{(1 - 5 x) \cdot 4 + (1 - 5 x) \cdot (5 (\frac{3 + 4 x}{1 - 5 x}))}$

Add $4 x$ and $15 x$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) \cdot 4 + (1 - 5 x) \cdot (5 (\frac{3 + 4 x}{1 - 5 x}))}$

Simplify the denominator.

Factor $1 - 5 x$ out of $(1 - 5 x) \cdot 4 + (1 - 5 x) \cdot 5 \frac{3 + 4 x}{1 - 5 x}$ .

Factor $1 - 5 x$ out of $(1 - 5 x) \cdot 4$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) (4) + (1 - 5 x) (5 (\frac{3 + 4 x}{1 - 5 x}))}$

Factor $1 - 5 x$ out of $(1 - 5 x) (4) + (1 - 5 x) (5 \frac{3 + 4 x}{1 - 5 x})$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) (4 + 5 (\frac{3 + 4 x}{1 - 5 x}))}$

Combine $5$ and $\frac{3 + 4 x}{1 - 5 x}$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) (4 + \frac{5 (3 + 4 x)}{1 - 5 x})}$

To write $4$ as a fraction with a common denominator, multiply by $\frac{1 - 5 x}{1 - 5 x}$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) (\frac{4 (1 - 5 x)}{1 - 5 x} + \frac{5 (3 + 4 x)}{1 - 5 x})}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) (\frac{4 (1 - 5 x) + 5 (3 + 4 x)}{1 - 5 x})}$

Rewrite $\frac{4 (1 - 5 x) + 5 (3 + 4 x)}{1 - 5 x}$ in a factored form.

Apply the distributive property.

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) (\frac{4 \cdot 1 + 4 (- 5 x) + 5 (3 + 4 x)}{1 - 5 x})}$

Multiply $4$ by $1$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) (\frac{4 + 4 (- 5 x) + 5 (3 + 4 x)}{1 - 5 x})}$

Multiply $- 5$ by $4$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) (\frac{4 - 20 x + 5 (3 + 4 x)}{1 - 5 x})}$

Apply the distributive property.

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) (\frac{4 - 20 x + 5 \cdot 3 + 5 (4 x)}{1 - 5 x})}$

Multiply $5$ by $3$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) (\frac{4 - 20 x + 15 + 5 (4 x)}{1 - 5 x})}$

Multiply $4$ by $5$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) (\frac{4 - 20 x + 15 + 20 x}{1 - 5 x})}$

Add $4$ and $15$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) (\frac{- 20 x + 19 + 20 x}{1 - 5 x})}$

Add $- 20 x$ and $20 x$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) (\frac{0 + 19}{1 - 5 x})}$

Add $0$ and $19$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{19 x}{(1 - 5 x) (\frac{19}{1 - 5 x})}$

Factor $\frac{19}{1 - 5 x}$ out of $\frac{19 x}{(1 - 5 x) \frac{19}{1 - 5 x}}$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{1 - 5 x}{19} \cdot \frac{19 x}{1 - 5 x}$

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

Cancel the common factor.

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{1 - 5 x}{19} \cdot \frac{19 x}{1 - 5 x}$

Rewrite the expression.

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{1}{19} \cdot (19 x)$

Cancel the common factor of $19$ .

Factor $19$ out of $19 x$ .

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{1}{19} \cdot (19 (x))$

Cancel the common factor.

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = \frac{1}{19} \cdot (19 x)$

Rewrite the expression.

$f^{- 1} (\frac{3 + 4 x}{1 - 5 x}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{x - 3}{4 + 5 x})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{3 + 4 (\frac{x - 3}{4 + 5 x})}{1 - 5 \frac{x - 3}{4 + 5 x}}$

Simplify the numerator.

Combine $4$ and $\frac{x - 3}{4 + 5 x}$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{3 + \frac{4 (x - 3)}{4 + 5 x}}{1 - 5 \frac{x - 3}{4 + 5 x}}$

To write $3$ as a fraction with a common denominator, multiply by $\frac{4 + 5 x}{4 + 5 x}$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{3 (4 + 5 x)}{4 + 5 x} + \frac{4 (x - 3)}{4 + 5 x}}{1 - 5 \frac{x - 3}{4 + 5 x}}$

Combine the numerators over the common denominator.

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{3 (4 + 5 x) + 4 (x - 3)}{4 + 5 x}}{1 - 5 \frac{x - 3}{4 + 5 x}}$

Rewrite $\frac{3 (4 + 5 x) + 4 (x - 3)}{4 + 5 x}$ in a factored form.

Apply the distributive property.

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{3 \cdot 4 + 3 (5 x) + 4 (x - 3)}{4 + 5 x}}{1 - 5 \frac{x - 3}{4 + 5 x}}$

Multiply $3$ by $4$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{12 + 3 (5 x) + 4 (x - 3)}{4 + 5 x}}{1 - 5 \frac{x - 3}{4 + 5 x}}$

Multiply $5$ by $3$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{12 + 15 x + 4 (x - 3)}{4 + 5 x}}{1 - 5 \frac{x - 3}{4 + 5 x}}$

Apply the distributive property.

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{12 + 15 x + 4 x + 4 \cdot - 3}{4 + 5 x}}{1 - 5 \frac{x - 3}{4 + 5 x}}$

Multiply $4$ by $- 3$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{12 + 15 x + 4 x - 12}{4 + 5 x}}{1 - 5 \frac{x - 3}{4 + 5 x}}$

Subtract $12$ from $12$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{15 x + 4 x + 0}{4 + 5 x}}{1 - 5 \frac{x - 3}{4 + 5 x}}$

Add $15 x + 4 x$ and $0$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{15 x + 4 x}{4 + 5 x}}{1 - 5 \frac{x - 3}{4 + 5 x}}$

Add $15 x$ and $4 x$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{19 x}{4 + 5 x}}{1 - 5 \frac{x - 3}{4 + 5 x}}$

Simplify the denominator.

Combine $- 5$ and $\frac{x - 3}{4 + 5 x}$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{19 x}{4 + 5 x}}{1 + \frac{- 5 (x - 3)}{4 + 5 x}}$

Move the negative in front of the fraction.

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{19 x}{4 + 5 x}}{1 - \frac{(5) (x - 3)}{4 + 5 x}}$

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

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{19 x}{4 + 5 x}}{\frac{4 + 5 x}{4 + 5 x} - \frac{5 (x - 3)}{4 + 5 x}}$

Combine the numerators over the common denominator.

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{19 x}{4 + 5 x}}{\frac{4 + 5 x - 5 (x - 3)}{4 + 5 x}}$

Reorder terms.

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{19 x}{4 + 5 x}}{\frac{5 x + 4 - 5 (x - 3)}{5 x + 4}}$

Rewrite $\frac{5 x + 4 - 5 (x - 3)}{5 x + 4}$ in a factored form.

Apply the distributive property.

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{19 x}{4 + 5 x}}{\frac{5 x + 4 - 5 x - 5 \cdot - 3}{5 x + 4}}$

Multiply $- 5$ by $- 3$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{19 x}{4 + 5 x}}{\frac{5 x + 4 - 5 x + 15}{5 x + 4}}$

Subtract $5 x$ from $5 x$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{19 x}{4 + 5 x}}{\frac{0 + 4 + 15}{5 x + 4}}$

Add $0$ and $4$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{19 x}{4 + 5 x}}{\frac{4 + 15}{5 x + 4}}$

Add $4$ and $15$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{\frac{19 x}{4 + 5 x}}{\frac{19}{5 x + 4}}$

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{x - 3}{4 + 5 x}) = \frac{19 x}{4 + 5 x} \cdot \frac{5 x + 4}{19}$

Cancel the common factor of $19$ .

Factor $19$ out of $19 x$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{19 (x)}{4 + 5 x} \cdot \frac{5 x + 4}{19}$

Cancel the common factor.

$f (\frac{x - 3}{4 + 5 x}) = \frac{19 x}{4 + 5 x} \cdot \frac{5 x + 4}{19}$

Rewrite the expression.

$f (\frac{x - 3}{4 + 5 x}) = \frac{x}{4 + 5 x} \cdot (5 x + 4)$

Multiply $\frac{x}{4 + 5 x}$ by $5 x + 4$ .

$f (\frac{x - 3}{4 + 5 x}) = \frac{x (5 x + 4)}{4 + 5 x}$

Cancel the common factor of $5 x + 4$ and $4 + 5 x$ .

Reorder terms.

$f (\frac{x - 3}{4 + 5 x}) = \frac{x (5 x + 4)}{5 x + 4}$

Cancel the common factor.

$f (\frac{x - 3}{4 + 5 x}) = \frac{x (5 x + 4)}{5 x + 4}$

Divide $x$ by $1$ .

$f (\frac{x - 3}{4 + 5 x}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x - 3}{4 + 5 x}$ is the inverse of $f (x) = \frac{3 + 4 x}{1 - 5 x}$ .

$f^{- 1} (x) = \frac{x - 3}{4 + 5 x}$

Find the Inverse (3+4x)/(1-5x)

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