Algebra Examples

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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Step 2.1

Multiply the equation by $y - 1$ .

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

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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Step 2.2.1.1

Apply the distributive property.

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

Step 2.2.1.2

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

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

Step 2.3

Simplify the right side.

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Step 2.3.1

Cancel the common factor of $y - 1$ .

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Step 2.3.1.1

Cancel the common factor.

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

Step 2.3.1.2

Rewrite the expression.

$y x - x = 2 y + 3$

Step 2.4

Solve for $y$ .

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Step 2.4.1

Subtract $2 y$ from both sides of the equation.

$y x - x - 2 y = 3$

Step 2.4.2

Add $x$ to both sides of the equation.

$y x - 2 y = 3 + x$

Step 2.4.3

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

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Step 2.4.3.1

Factor $y$ out of $y x$ .

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

Step 2.4.3.2

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

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

Step 2.4.3.3

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

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

Step 2.4.4

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

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Step 2.4.4.1

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

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

Step 2.4.4.2

Simplify the left side.

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Step 2.4.4.2.1

Cancel the common factor of $x - 2$ .

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Step 2.4.4.2.1.1

Cancel the common factor.

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

Step 2.4.4.2.1.2

Divide $y$ by $1$ .

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

Step 2.4.4.3

Simplify the right side.

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Step 2.4.4.3.1

Combine the numerators over the common denominator.

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

Step 3

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

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

Step 4

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

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Step 4.1

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Step 4.2

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

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

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

Step 4.2.3

Remove parentheses.

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

Step 4.2.4

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

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Step 4.2.4.1

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

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

Step 4.2.4.2

Combine.

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

Step 4.2.5

Apply the distributive property.

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

Step 4.2.6

Simplify by cancelling.

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Step 4.2.6.1

Cancel the common factor of $x - 1$ .

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Step 4.2.6.1.1

Cancel the common factor.

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

Step 4.2.6.1.2

Rewrite the expression.

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

Step 4.2.6.2

Cancel the common factor of $x - 1$ .

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Step 4.2.6.2.1

Cancel the common factor.

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

Step 4.2.6.2.2

Rewrite the expression.

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

Step 4.2.7

Simplify the numerator.

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Step 4.2.7.1

Apply the distributive property.

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

Step 4.2.7.2

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

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

Step 4.2.7.3

Multiply $- 1$ by $3$ .

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

Step 4.2.7.4

Add $3 x$ and $2 x$ .

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

Step 4.2.7.5

Add $- 3$ and $3$ .

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

Step 4.2.7.6

Add $5 x$ and $0$ .

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

Step 4.2.8

Simplify the denominator.

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Step 4.2.8.1

Apply the distributive property.

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

Step 4.2.8.2

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

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

Step 4.2.8.3

Multiply $- 1$ by $- 2$ .

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

Step 4.2.8.4

Subtract $2 x$ from $2 x$ .

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

Step 4.2.8.5

Add $0$ and $3$ .

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

Step 4.2.8.6

Add $3$ and $2$ .

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

Step 4.2.9

Cancel the common factor of $5$ .

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Step 4.2.9.1

Cancel the common factor.

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

Step 4.2.9.2

Divide $x$ by $1$ .

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

Step 4.3

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

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

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

Step 4.3.3

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

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Step 4.3.3.1

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

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

Step 4.3.3.2

Combine.

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

Step 4.3.4

Apply the distributive property.

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

Step 4.3.5

Cancel the common factor of $x - 2$ .

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Step 4.3.5.1

Cancel the common factor.

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

Step 4.3.5.2

Rewrite the expression.

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

Step 4.3.6

Simplify the numerator.

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Step 4.3.6.1

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

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Step 4.3.6.1.1

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

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

Step 4.3.6.1.2

Factor $x - 2$ out of $(x - 2) \cdot 3$ .

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

Step 4.3.6.1.3

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

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

Step 4.3.6.2

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

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

Step 4.3.6.3

To write $3$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

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

Step 4.3.6.4

Combine the numerators over the common denominator.

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

Step 4.3.6.5

Reorder terms.

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

Step 4.3.6.6

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

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Step 4.3.6.6.1

Apply the distributive property.

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

Step 4.3.6.6.2

Multiply $3$ by $- 2$ .

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

Step 4.3.6.6.3

Apply the distributive property.

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

Step 4.3.6.6.4

Multiply $2$ by $3$ .

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

Step 4.3.6.6.5

Add $3 x$ and $2 x$ .

$f (\frac{3 + x}{x - 2}) = \frac{(x - 2) (\frac{5 x - 6 + 6}{x - 2})}{3 + x + (x - 2) \cdot - 1}$

Step 4.3.6.6.6

Add $- 6$ and $6$ .

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

Step 4.3.6.6.7

Add $5 x$ and $0$ .

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

Step 4.3.7

Simplify the denominator.

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Step 4.3.7.1

Apply the distributive property.

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

Step 4.3.7.2

Move $- 1$ to the left of $x$ .

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

Step 4.3.7.3

Multiply $- 2$ by $- 1$ .

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

Step 4.3.7.4

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

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

Step 4.3.7.5

Add $3$ and $2$ .

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

Step 4.3.7.6

Subtract $x$ from $x$ .

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

Step 4.3.7.7

Add $0$ and $5$ .

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

Step 4.3.8

Reduce the expression by cancelling the common factors.

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Step 4.3.8.1

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

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

Step 4.3.8.2

Cancel the common factor of $5$ .

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Step 4.3.8.2.1

Factor $5$ out of $5 x$ .

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

Step 4.3.8.2.2

Cancel the common factor.

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

Step 4.3.8.2.3

Rewrite the expression.

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

Step 4.3.8.3

Cancel the common factor of $x - 2$ .

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Step 4.3.8.3.1

Cancel the common factor.

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

Step 4.3.8.3.2

Rewrite the expression.

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

Step 4.4

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

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