Find the Inverse F(x)=(2x-1)/(x+3)

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

Step 1

Write $F (x) = \frac{2 x - 1}{x + 3}$ as an equation.

$y = \frac{2 x - 1}{x + 3}$

Step 2

Interchange the variables.

$x = \frac{2 y - 1}{y + 3}$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{2 y - 1}{y + 3} = x$ .

$\frac{2 y - 1}{y + 3} = x$

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.

$y + 3, 1$

Remove parentheses.

$y + 3, 1$

The LCM of one and any expression is the expression.

$y + 3$

Multiply each term in $\frac{2 y - 1}{y + 3} = x$ by $y + 3$ to eliminate the fractions.

Multiply each term in $\frac{2 y - 1}{y + 3} = x$ by $y + 3$ .

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

Simplify the left side.

Cancel the common factor of $y + 3$ .

Cancel the common factor.

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

Rewrite the expression.

$2 y - 1 = x (y + 3)$

Simplify the right side.

Apply the distributive property.

$2 y - 1 = x y + x \cdot 3$

Move $3$ to the left of $x$ .

$2 y - 1 = x y + 3 x$

Solve the equation.

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

$2 y - 1 - x y = 3 x$

Add $1$ to both sides of the equation.

$2 y - x y = 3 x + 1$

Factor $y$ out of $2 y - x y$ .

Factor $y$ out of $2 y$ .

$y \cdot 2 - x y = 3 x + 1$

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

$y \cdot 2 + y (- x) = 3 x + 1$

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

$y (2 - x) = 3 x + 1$

Divide each term in $y (2 - x) = 3 x + 1$ by $2 - x$ and simplify.

Divide each term in $y (2 - x) = 3 x + 1$ by $2 - x$ .

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

Simplify the left side.

Cancel the common factor of $2 - x$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

$y = \frac{3 x}{2 - x} + \frac{1}{2 - x}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{3 x + 1}{2 - x}$

Step 4

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

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

Step 5

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

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

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

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

Multiply $\frac{3 \frac{2 x - 1}{x + 3} + 1}{2 - \frac{2 x - 1}{x + 3}}$ by $\frac{x + 3}{x + 3}$ .

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

Combine.

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

Apply the distributive property.

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

Cancel the common factor of $x + 3$ .

Move the leading negative in $- \frac{2 x - 1}{x + 3}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Factor $x + 3$ out of $(x + 3) \cdot 3 \frac{2 x - 1}{x + 3} + (x + 3) \cdot 1$ .

Factor $x + 3$ out of $(x + 3) \cdot 3 \frac{2 x - 1}{x + 3}$ .

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

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

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

Factor $x + 3$ out of $(x + 3) (3 \frac{2 x - 1}{x + 3}) + (x + 3) (1)$ .

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

Combine $3$ and $\frac{2 x - 1}{x + 3}$ .

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

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

Rewrite $\frac{x + 3 (2 x - 1) + 3}{x + 3}$ in a factored form.

Apply the distributive property.

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

Multiply $2$ by $3$ .

$F^{- 1} (\frac{2 x - 1}{x + 3}) = \frac{(x + 3) (\frac{x + 6 x + 3 \cdot - 1 + 3}{x + 3})}{(x + 3) \cdot 2 - (2 x - 1)}$

Multiply $3$ by $- 1$ .

$F^{- 1} (\frac{2 x - 1}{x + 3}) = \frac{(x + 3) (\frac{x + 6 x - 3 + 3}{x + 3})}{(x + 3) \cdot 2 - (2 x - 1)}$

Add $x$ and $6 x$ .

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

Add $- 3$ and $3$ .

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

Add $7 x$ and $0$ .

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

Simplify the denominator.

Apply the distributive property.

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

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

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

Multiply $3$ by $2$ .

$F^{- 1} (\frac{2 x - 1}{x + 3}) = \frac{(x + 3) (\frac{7 x}{x + 3})}{2 \cdot x + 6 - (2 x - 1)}$

Apply the distributive property.

$F^{- 1} (\frac{2 x - 1}{x + 3}) = \frac{(x + 3) (\frac{7 x}{x + 3})}{2 x + 6 - (2 x) + 1}$

Multiply $2$ by $- 1$ .

$F^{- 1} (\frac{2 x - 1}{x + 3}) = \frac{(x + 3) (\frac{7 x}{x + 3})}{2 x + 6 - 2 x + 1}$

Multiply $- 1$ by $- 1$ .

$F^{- 1} (\frac{2 x - 1}{x + 3}) = \frac{(x + 3) (\frac{7 x}{x + 3})}{2 x + 6 - 2 x + 1}$

Subtract $2 x$ from $2 x$ .

$F^{- 1} (\frac{2 x - 1}{x + 3}) = \frac{(x + 3) (\frac{7 x}{x + 3})}{0 + 6 + 1}$

Add $0$ and $6$ .

$F^{- 1} (\frac{2 x - 1}{x + 3}) = \frac{(x + 3) (\frac{7 x}{x + 3})}{6 + 1}$

Add $6$ and $1$ .

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

Reduce the expression by cancelling the common factors.

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

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

Cancel the common factor of $7$ .

Factor $7$ out of $7 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Set up the composite result function.

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

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

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

Multiply the numerator and denominator of the complex fraction by $2 - x$ .

Multiply $\frac{2 \frac{3 x + 1}{2 - x} - 1}{\frac{3 x + 1}{2 - x} + 3}$ by $\frac{2 - x}{2 - x}$ .

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

Combine.

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

Apply the distributive property.

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

Cancel the common factor of $2 - x$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Factor $2 - x$ out of $(2 - x) \cdot 2 \frac{3 x + 1}{2 - x} + (2 - x) \cdot - 1$ .

Factor $2 - x$ out of $(2 - x) \cdot 2 \frac{3 x + 1}{2 - x}$ .

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

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

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

Factor $2 - x$ out of $(2 - x) (2 \frac{3 x + 1}{2 - x}) + (2 - x) (- 1)$ .

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

Combine $2$ and $\frac{3 x + 1}{2 - x}$ .

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

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

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

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

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

Combine the numerators over the common denominator.

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

Reorder terms.

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

Rewrite $\frac{2 (3 x + 1) - (2 - x)}{- x + 2}$ in a factored form.

Apply the distributive property.

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

Multiply $3$ by $2$ .

$F (\frac{3 x + 1}{2 - x}) = \frac{(2 - x) (\frac{6 x + 2 \cdot 1 - (2 - x)}{- x + 2})}{3 x + 1 + (2 - x) \cdot 3}$

Multiply $2$ by $1$ .

$F (\frac{3 x + 1}{2 - x}) = \frac{(2 - x) (\frac{6 x + 2 - (2 - x)}{- x + 2})}{3 x + 1 + (2 - x) \cdot 3}$

Apply the distributive property.

$F (\frac{3 x + 1}{2 - x}) = \frac{(2 - x) (\frac{6 x + 2 - 1 \cdot 2 + x}{- x + 2})}{3 x + 1 + (2 - x) \cdot 3}$

Multiply $- 1$ by $2$ .

$F (\frac{3 x + 1}{2 - x}) = \frac{(2 - x) (\frac{6 x + 2 - 2 + x}{- x + 2})}{3 x + 1 + (2 - x) \cdot 3}$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

$F (\frac{3 x + 1}{2 - x}) = \frac{(2 - x) (\frac{6 x + 2 - 2 + 1 x}{- x + 2})}{3 x + 1 + (2 - x) \cdot 3}$

Multiply $x$ by $1$ .

$F (\frac{3 x + 1}{2 - x}) = \frac{(2 - x) (\frac{6 x + 2 - 2 + x}{- x + 2})}{3 x + 1 + (2 - x) \cdot 3}$

Add $6 x$ and $x$ .

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

Subtract $2$ from $2$ .

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

Add $7 x$ and $0$ .

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

Simplify the denominator.

Apply the distributive property.

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

Multiply $2$ by $3$ .

$F (\frac{3 x + 1}{2 - x}) = \frac{(2 - x) (\frac{7 x}{- x + 2})}{3 x + 1 + 6 - x \cdot 3}$

Multiply $3$ by $- 1$ .

$F (\frac{3 x + 1}{2 - x}) = \frac{(2 - x) (\frac{7 x}{- x + 2})}{3 x + 1 + 6 - 3 x}$

Subtract $3 x$ from $3 x$ .

$F (\frac{3 x + 1}{2 - x}) = \frac{(2 - x) (\frac{7 x}{- x + 2})}{0 + 1 + 6}$

Add $0$ and $1$ .

$F (\frac{3 x + 1}{2 - x}) = \frac{(2 - x) (\frac{7 x}{- x + 2})}{1 + 6}$

Add $1$ and $6$ .

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

Simplify terms.

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

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

Cancel the common factor of $7$ .

Factor $7$ out of $7 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Cancel the common factor of $2 - x$ and $- x + 2$ .

Reorder terms.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse F(x)=(2x-1)/(x+3)

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