Algebra Examples

$\frac{x + 2}{x - 5}$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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Step 2.1

Multiply the equation by $y - 5$ .

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

Step 2.2

Simplify the left side.

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Step 2.2.1

Apply the distributive property.

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

Step 2.3

Simplify the right side.

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Step 2.3.1

Cancel the common factor of $y - 5$ .

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Step 2.3.1.1

Cancel the common factor.

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

Step 2.3.1.2

Rewrite the expression.

$y x - 5 x = y + 2$

Step 2.4

Solve for $y$ .

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Step 2.4.1

Subtract $y$ from both sides of the equation.

$y x - 5 x - y = 2$

Step 2.4.2

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

$y x - y = 2 + 5 x$

Step 2.4.3

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

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Step 2.4.3.1

Factor $y$ out of $y x$ .

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

Step 2.4.3.2

Factor $y$ out of $- y$ .

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

Step 2.4.3.3

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

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

Step 2.4.4

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

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Step 2.4.4.1

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

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

Step 2.4.4.2

Simplify the left side.

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Step 2.4.4.2.1

Cancel the common factor of $x - 1$ .

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Step 2.4.4.2.1.1

Cancel the common factor.

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

Step 2.4.4.2.1.2

Divide $y$ by $1$ .

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

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.

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

Step 3

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

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

Step 4

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

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

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

Step 4.2.3

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

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Step 4.2.3.1

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

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

Step 4.2.3.2

Combine.

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

Step 4.2.4

Apply the distributive property.

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

Step 4.2.5

Cancel the common factor of $x - 5$ .

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Step 4.2.5.1

Cancel the common factor.

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

Step 4.2.5.2

Rewrite the expression.

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

Step 4.2.6

Simplify the numerator.

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Step 4.2.6.1

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

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Step 4.2.6.1.1

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

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

Step 4.2.6.1.2

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

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

Step 4.2.6.2

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

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

Step 4.2.6.3

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

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

Step 4.2.6.4

Combine the numerators over the common denominator.

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

Step 4.2.6.5

Reorder terms.

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

Step 4.2.6.6

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

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Step 4.2.6.6.1

Apply the distributive property.

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

Step 4.2.6.6.2

Multiply $5$ by $2$ .

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

Step 4.2.6.6.3

Apply the distributive property.

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

Step 4.2.6.6.4

Multiply $2$ by $- 5$ .

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

Step 4.2.6.6.5

Add $5 x$ and $2 x$ .

$f^{- 1} (\frac{x + 2}{x - 5}) = \frac{(x - 5) (\frac{7 x + 10 - 10}{x - 5})}{x + 2 + (x - 5) \cdot - 1}$

Step 4.2.6.6.6

Subtract $10$ from $10$ .

$f^{- 1} (\frac{x + 2}{x - 5}) = \frac{(x - 5) (\frac{7 x + 0}{x - 5})}{x + 2 + (x - 5) \cdot - 1}$

Step 4.2.6.6.7

Add $7 x$ and $0$ .

$f^{- 1} (\frac{x + 2}{x - 5}) = \frac{(x - 5) (\frac{7 x}{x - 5})}{x + 2 + (x - 5) \cdot - 1}$

Step 4.2.7

Simplify the denominator.

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Step 4.2.7.1

Apply the distributive property.

$f^{- 1} (\frac{x + 2}{x - 5}) = \frac{(x - 5) (\frac{7 x}{x - 5})}{x + 2 + x \cdot - 1 - 5 \cdot - 1}$

Step 4.2.7.2

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

$f^{- 1} (\frac{x + 2}{x - 5}) = \frac{(x - 5) (\frac{7 x}{x - 5})}{x + 2 - 1 \cdot x - 5 \cdot - 1}$

Step 4.2.7.3

Multiply $- 5$ by $- 1$ .

$f^{- 1} (\frac{x + 2}{x - 5}) = \frac{(x - 5) (\frac{7 x}{x - 5})}{x + 2 - 1 \cdot x + 5}$

Step 4.2.7.4

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

$f^{- 1} (\frac{x + 2}{x - 5}) = \frac{(x - 5) (\frac{7 x}{x - 5})}{x + 2 - x + 5}$

Step 4.2.7.5

Subtract $x$ from $x$ .

$f^{- 1} (\frac{x + 2}{x - 5}) = \frac{(x - 5) (\frac{7 x}{x - 5})}{0 + 2 + 5}$

Step 4.2.7.6

Add $0$ and $2$ .

$f^{- 1} (\frac{x + 2}{x - 5}) = \frac{(x - 5) (\frac{7 x}{x - 5})}{2 + 5}$

Step 4.2.7.7

Add $2$ and $5$ .

$f^{- 1} (\frac{x + 2}{x - 5}) = \frac{(x - 5) (\frac{7 x}{x - 5})}{7}$

Step 4.2.8

Reduce the expression by cancelling the common factors.

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Step 4.2.8.1

Factor $\frac{7 x}{x - 5}$ out of $\frac{(x - 5) \frac{7 x}{x - 5}}{7}$ .

$f^{- 1} (\frac{x + 2}{x - 5}) = \frac{7 x}{x - 5} \cdot \frac{x - 5}{7}$

Step 4.2.8.2

Cancel the common factor of $7$ .

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Step 4.2.8.2.1

Factor $7$ out of $7 x$ .

$f^{- 1} (\frac{x + 2}{x - 5}) = \frac{7 (x)}{x - 5} \cdot \frac{x - 5}{7}$

Step 4.2.8.2.2

Cancel the common factor.

$f^{- 1} (\frac{x + 2}{x - 5}) = \frac{7 x}{x - 5} \cdot \frac{x - 5}{7}$

Step 4.2.8.2.3

Rewrite the expression.

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

Step 4.2.8.3

Cancel the common factor of $x - 5$ .

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Step 4.2.8.3.1

Cancel the common factor.

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

Step 4.2.8.3.2

Rewrite the expression.

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

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

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

Step 4.3.3

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

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Step 4.3.3.1

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

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

Step 4.3.3.2

Combine.

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

Step 4.3.4

Apply the distributive property.

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

Step 4.3.5

Simplify by cancelling.

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Step 4.3.5.1

Cancel the common factor of $x - 1$ .

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Step 4.3.5.1.1

Cancel the common factor.

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

Step 4.3.5.1.2

Rewrite the expression.

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

Step 4.3.5.2

Cancel the common factor of $x - 1$ .

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Step 4.3.5.2.1

Cancel the common factor.

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

Step 4.3.5.2.2

Rewrite the expression.

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

Step 4.3.6

Simplify the numerator.

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Step 4.3.6.1

Apply the distributive property.

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

Step 4.3.6.2

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

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

Step 4.3.6.3

Multiply $- 1$ by $2$ .

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

Step 4.3.6.4

Subtract $2$ from $2$ .

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

Step 4.3.6.5

Add $5 x + 2 x$ and $0$ .

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

Step 4.3.6.6

Add $5 x$ and $2 x$ .

$f (\frac{2 + 5 x}{x - 1}) = \frac{7 x}{2 + 5 x + (x - 1) \cdot - 5}$

Step 4.3.7

Simplify the denominator.

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Step 4.3.7.1

Apply the distributive property.

$f (\frac{2 + 5 x}{x - 1}) = \frac{7 x}{2 + 5 x + x \cdot - 5 - 1 \cdot - 5}$

Step 4.3.7.2

Move $- 5$ to the left of $x$ .

$f (\frac{2 + 5 x}{x - 1}) = \frac{7 x}{2 + 5 x - 5 \cdot x - 1 \cdot - 5}$

Step 4.3.7.3

Multiply $- 1$ by $- 5$ .

$f (\frac{2 + 5 x}{x - 1}) = \frac{7 x}{2 + 5 x - 5 \cdot x + 5}$

Step 4.3.7.4

Add $2$ and $5$ .

$f (\frac{2 + 5 x}{x - 1}) = \frac{7 x}{5 x - 5 x + 7}$

Step 4.3.7.5

Subtract $5 x$ from $5 x$ .

$f (\frac{2 + 5 x}{x - 1}) = \frac{7 x}{0 + 7}$

Step 4.3.7.6

Add $0$ and $7$ .

$f (\frac{2 + 5 x}{x - 1}) = \frac{7 x}{7}$

Step 4.3.8

Cancel the common factor of $7$ .

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Step 4.3.8.1

Cancel the common factor.

$f (\frac{2 + 5 x}{x - 1}) = \frac{7 x}{7}$

Step 4.3.8.2

Divide $x$ by $1$ .

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

Step 4.4

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

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