Find the Inverse y=x/(3x+1)

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

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.

$3 y + 1, 1$

Remove parentheses.

$3 y + 1, 1$

The LCM of one and any expression is the expression.

$3 y + 1$

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

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

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

Simplify the left side.

Cancel the common factor of $3 y + 1$ .

Cancel the common factor.

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

Rewrite the expression.

$y = x (3 y + 1)$

Simplify the right side.

Apply the distributive property.

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

Simplify the expression.

Rewrite using the commutative property of multiplication.

$y = 3 x y + x \cdot 1$

Multiply $x$ by $1$ .

$y = 3 x y + x$

Solve the equation.

Subtract $3 x y$ from both sides of the equation.

$y - 3 x y = x$

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

Factor $y$ out of $y^{1}$ .

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

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

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

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

$y (1 - 3 x) = x$

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

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

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

Simplify the left side.

Cancel the common factor of $1 - 3 x$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Step 3

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

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

Step 4

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

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

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

Multiply the numerator by the reciprocal of the denominator.

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

Simplify the denominator.

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

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

Move the negative in front of the fraction.

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

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

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

Subtract $3 x$ from $3 x$ .

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

Add $0$ and $1$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $3 x + 1$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Set up the composite result function.

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

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

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

Multiply the numerator by the reciprocal of the denominator.

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

Simplify the denominator.

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

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

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

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

Subtract $3 x$ from $3 x$ .

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

Add $0$ and $1$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $- 3 x + 1$ by $1$ .

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

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

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

Cancel the common factor of $- 3 x + 1$ and $1 - 3 x$ .

Reorder terms.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse y=x/(3x+1)

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