Precalculus Examples

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

Step 1

Write $f (x) = \frac{2 x - 3}{5 x + 1}$ as an equation.

$y = \frac{2 x - 3}{5 x + 1}$

Step 2

Interchange the variables.

$x = \frac{2 y - 3}{5 y + 1}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{2 y - 3}{5 y + 1} = x$ .

$\frac{2 y - 3}{5 y + 1} = x$

Step 3.2

Find the LCD of the terms in the equation.

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Step 3.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$5 y + 1, 1$

Step 3.2.2

Remove parentheses.

$5 y + 1, 1$

Step 3.2.3

The LCM of one and any expression is the expression.

$5 y + 1$

Step 3.3

Multiply each term in $\frac{2 y - 3}{5 y + 1} = x$ by $5 y + 1$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $\frac{2 y - 3}{5 y + 1} = x$ by $5 y + 1$ .

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

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

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Step 3.3.2.1.1

Cancel the common factor.

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

Step 3.3.2.1.2

Rewrite the expression.

$2 y - 3 = x (5 y + 1)$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Apply the distributive property.

$2 y - 3 = x (5 y) + x \cdot 1$

Step 3.3.3.2

Simplify the expression.

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Step 3.3.3.2.1

Rewrite using the commutative property of multiplication.

$2 y - 3 = 5 x y + x \cdot 1$

Step 3.3.3.2.2

Multiply $x$ by $1$ .

$2 y - 3 = 5 x y + x$

Step 3.4

Solve the equation.

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Step 3.4.1

Subtract $5 x y$ from both sides of the equation.

$2 y - 3 - 5 x y = x$

Step 3.4.2

Add $3$ to both sides of the equation.

$2 y - 5 x y = x + 3$

Step 3.4.3

Factor $y$ out of $2 y - 5 x y$ .

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Step 3.4.3.1

Factor $y$ out of $2 y$ .

$y \cdot 2 - 5 x y = x + 3$

Step 3.4.3.2

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

$y \cdot 2 + y (- 5 x) = x + 3$

Step 3.4.3.3

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

$y (2 - 5 x) = x + 3$

Step 3.4.4

Divide each term in $y (2 - 5 x) = x + 3$ by $2 - 5 x$ and simplify.

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Step 3.4.4.1

Divide each term in $y (2 - 5 x) = x + 3$ by $2 - 5 x$ .

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

Step 3.4.4.2

Simplify the left side.

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Step 3.4.4.2.1

Cancel the common factor of $2 - 5 x$ .

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Step 3.4.4.2.1.1

Cancel the common factor.

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

Step 3.4.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{x}{2 - 5 x} + \frac{3}{2 - 5 x}$

Step 3.4.4.3

Simplify the right side.

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Step 3.4.4.3.1

Combine the numerators over the common denominator.

$y = \frac{x + 3}{2 - 5 x}$

Step 4

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

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

Step 5

Verify if $f^{- 1} (x) = \frac{x + 3}{2 - 5 x}$ is the inverse of $f (x) = \frac{2 x - 3}{5 x + 1}$ .

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Step 5.1

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

Step 5.2

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

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

Evaluate $f^{- 1} (\frac{2 x - 3}{5 x + 1})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Step 5.2.3

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

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Step 5.2.3.1

Multiply $\frac{\frac{2 x - 3}{5 x + 1} + 3}{2 - 5 \frac{2 x - 3}{5 x + 1}}$ by $\frac{5 x + 1}{5 x + 1}$ .

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

Step 5.2.3.2

Combine.

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

Step 5.2.4

Apply the distributive property.

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

Step 5.2.5

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

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Step 5.2.5.1

Cancel the common factor.

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

Step 5.2.5.2

Rewrite the expression.

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

Step 5.2.6

Simplify the numerator.

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Step 5.2.6.1

Apply the distributive property.

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

Step 5.2.6.2

Multiply $3$ by $5$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{2 x - 3 + 15 x + 1 \cdot 3}{(5 x + 1) \cdot 2 + (5 x + 1) (- 5 \frac{2 x - 3}{5 x + 1})}$

Step 5.2.6.3

Multiply $3$ by $1$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{2 x - 3 + 15 x + 3}{(5 x + 1) \cdot 2 + (5 x + 1) (- 5 \frac{2 x - 3}{5 x + 1})}$

Step 5.2.6.4

Add $2 x$ and $15 x$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x - 3 + 3}{(5 x + 1) \cdot 2 + (5 x + 1) (- 5 \frac{2 x - 3}{5 x + 1})}$

Step 5.2.6.5

Add $- 3$ and $3$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x + 0}{(5 x + 1) \cdot 2 + (5 x + 1) (- 5 \frac{2 x - 3}{5 x + 1})}$

Step 5.2.6.6

Add $17 x$ and $0$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) \cdot 2 + (5 x + 1) (- 5 \frac{2 x - 3}{5 x + 1})}$

Step 5.2.7

Simplify the denominator.

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Step 5.2.7.1

Factor $5 x + 1$ out of $(5 x + 1) \cdot 2 + (5 x + 1) (- 5 \frac{2 x - 3}{5 x + 1})$ .

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Step 5.2.7.1.1

Factor $5 x + 1$ out of $(5 x + 1) \cdot 2$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (2) + (5 x + 1) (- 5 \frac{2 x - 3}{5 x + 1})}$

Step 5.2.7.1.2

Factor $5 x + 1$ out of $(5 x + 1) (2) + (5 x + 1) (- 5 \frac{2 x - 3}{5 x + 1})$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (2 - 5 \frac{2 x - 3}{5 x + 1})}$

Step 5.2.7.2

Combine $- 5$ and $\frac{2 x - 3}{5 x + 1}$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (2 + \frac{- 5 (2 x - 3)}{5 x + 1})}$

Step 5.2.7.3

Move the negative in front of the fraction.

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (2 - \frac{(5) (2 x - 3)}{5 x + 1})}$

Step 5.2.7.4

To write $2$ as a fraction with a common denominator, multiply by $\frac{5 x + 1}{5 x + 1}$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (\frac{2 (5 x + 1)}{5 x + 1} - \frac{5 (2 x - 3)}{5 x + 1})}$

Step 5.2.7.5

Combine the numerators over the common denominator.

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (\frac{2 (5 x + 1) - 5 (2 x - 3)}{5 x + 1})}$

Step 5.2.7.6

Rewrite $\frac{2 (5 x + 1) - 5 (2 x - 3)}{5 x + 1}$ in a factored form.

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Step 5.2.7.6.1

Apply the distributive property.

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (\frac{2 (5 x) + 2 \cdot 1 - 5 (2 x - 3)}{5 x + 1})}$

Step 5.2.7.6.2

Multiply $5$ by $2$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (\frac{10 x + 2 \cdot 1 - 5 (2 x - 3)}{5 x + 1})}$

Step 5.2.7.6.3

Multiply $2$ by $1$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (\frac{10 x + 2 - 5 (2 x - 3)}{5 x + 1})}$

Step 5.2.7.6.4

Apply the distributive property.

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (\frac{10 x + 2 - 5 (2 x) - 5 \cdot - 3}{5 x + 1})}$

Step 5.2.7.6.5

Multiply $2$ by $- 5$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (\frac{10 x + 2 - 10 x - 5 \cdot - 3}{5 x + 1})}$

Step 5.2.7.6.6

Multiply $- 5$ by $- 3$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (\frac{10 x + 2 - 10 x + 15}{5 x + 1})}$

Step 5.2.7.6.7

Subtract $10 x$ from $10 x$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (\frac{0 + 2 + 15}{5 x + 1})}$

Step 5.2.7.6.8

Add $0$ and $2$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (\frac{2 + 15}{5 x + 1})}$

Step 5.2.7.6.9

Add $2$ and $15$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{17 x}{(5 x + 1) (\frac{17}{5 x + 1})}$

Step 5.2.8

Factor $\frac{17}{5 x + 1}$ out of $\frac{17 x}{(5 x + 1) \frac{17}{5 x + 1}}$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{5 x + 1}{17} \cdot \frac{17 x}{5 x + 1}$

Step 5.2.9

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

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Step 5.2.9.1

Cancel the common factor.

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{5 x + 1}{17} \cdot \frac{17 x}{5 x + 1}$

Step 5.2.9.2

Rewrite the expression.

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{1}{17} \cdot (17 x)$

Step 5.2.10

Cancel the common factor of $17$ .

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Step 5.2.10.1

Factor $17$ out of $17 x$ .

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{1}{17} \cdot (17 (x))$

Step 5.2.10.2

Cancel the common factor.

$f^{- 1} (\frac{2 x - 3}{5 x + 1}) = \frac{1}{17} \cdot (17 x)$

Step 5.2.10.3

Rewrite the expression.

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

Step 5.3

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

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

Evaluate $f (\frac{x + 3}{2 - 5 x})$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Step 5.3.3

Simplify the numerator.

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Step 5.3.3.1

Combine $2$ and $\frac{x + 3}{2 - 5 x}$ .

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

Step 5.3.3.2

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{2 - 5 x}{2 - 5 x}$ .

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

Step 5.3.3.3

Combine $- 3$ and $\frac{2 - 5 x}{2 - 5 x}$ .

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

Step 5.3.3.4

Combine the numerators over the common denominator.

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

Step 5.3.3.5

Reorder terms.

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

Step 5.3.3.6

Rewrite $\frac{2 (x + 3) - 3 (2 - 5 x)}{- 5 x + 2}$ in a factored form.

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Step 5.3.3.6.1

Apply the distributive property.

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

Step 5.3.3.6.2

Multiply $2$ by $3$ .

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

Step 5.3.3.6.3

Apply the distributive property.

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

Step 5.3.3.6.4

Multiply $- 3$ by $2$ .

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

Step 5.3.3.6.5

Multiply $- 5$ by $- 3$ .

$f (\frac{x + 3}{2 - 5 x}) = \frac{\frac{2 x + 6 - 6 + 15 x}{- 5 x + 2}}{5 (\frac{x + 3}{2 - 5 x}) + 1}$

Step 5.3.3.6.6

Add $2 x$ and $15 x$ .

$f (\frac{x + 3}{2 - 5 x}) = \frac{\frac{17 x + 6 - 6}{- 5 x + 2}}{5 (\frac{x + 3}{2 - 5 x}) + 1}$

Step 5.3.3.6.7

Subtract $6$ from $6$ .

$f (\frac{x + 3}{2 - 5 x}) = \frac{\frac{17 x + 0}{- 5 x + 2}}{5 (\frac{x + 3}{2 - 5 x}) + 1}$

Step 5.3.3.6.8

Add $17 x$ and $0$ .

$f (\frac{x + 3}{2 - 5 x}) = \frac{\frac{17 x}{- 5 x + 2}}{5 (\frac{x + 3}{2 - 5 x}) + 1}$

Step 5.3.4

Simplify the denominator.

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Step 5.3.4.1

Combine $5$ and $\frac{x + 3}{2 - 5 x}$ .

$f (\frac{x + 3}{2 - 5 x}) = \frac{\frac{17 x}{- 5 x + 2}}{\frac{5 (x + 3)}{2 - 5 x} + 1}$

Step 5.3.4.2

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

$f (\frac{x + 3}{2 - 5 x}) = \frac{\frac{17 x}{- 5 x + 2}}{\frac{5 (x + 3)}{2 - 5 x} + \frac{2 - 5 x}{2 - 5 x}}$

Step 5.3.4.3

Combine the numerators over the common denominator.

$f (\frac{x + 3}{2 - 5 x}) = \frac{\frac{17 x}{- 5 x + 2}}{\frac{5 (x + 3) + 2 - 5 x}{2 - 5 x}}$

Step 5.3.4.4

Reorder terms.

$f (\frac{x + 3}{2 - 5 x}) = \frac{\frac{17 x}{- 5 x + 2}}{\frac{- 5 x + 5 (x + 3) + 2}{- 5 x + 2}}$

Step 5.3.4.5

Rewrite $\frac{- 5 x + 5 (x + 3) + 2}{- 5 x + 2}$ in a factored form.

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Step 5.3.4.5.1

Apply the distributive property.

$f (\frac{x + 3}{2 - 5 x}) = \frac{\frac{17 x}{- 5 x + 2}}{\frac{- 5 x + 5 x + 5 \cdot 3 + 2}{- 5 x + 2}}$

Step 5.3.4.5.2

Multiply $5$ by $3$ .

$f (\frac{x + 3}{2 - 5 x}) = \frac{\frac{17 x}{- 5 x + 2}}{\frac{- 5 x + 5 x + 15 + 2}{- 5 x + 2}}$

Step 5.3.4.5.3

Add $- 5 x$ and $5 x$ .

$f (\frac{x + 3}{2 - 5 x}) = \frac{\frac{17 x}{- 5 x + 2}}{\frac{0 + 15 + 2}{- 5 x + 2}}$

Step 5.3.4.5.4

Add $0$ and $15$ .

$f (\frac{x + 3}{2 - 5 x}) = \frac{\frac{17 x}{- 5 x + 2}}{\frac{15 + 2}{- 5 x + 2}}$

Step 5.3.4.5.5

Add $15$ and $2$ .

$f (\frac{x + 3}{2 - 5 x}) = \frac{\frac{17 x}{- 5 x + 2}}{\frac{17}{- 5 x + 2}}$

Step 5.3.5

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{x + 3}{2 - 5 x}) = \frac{17 x}{- 5 x + 2} \cdot \frac{- 5 x + 2}{17}$

Step 5.3.6

Cancel the common factor of $17$ .

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Step 5.3.6.1

Factor $17$ out of $17 x$ .

$f (\frac{x + 3}{2 - 5 x}) = \frac{17 (x)}{- 5 x + 2} \cdot \frac{- 5 x + 2}{17}$

Step 5.3.6.2

Cancel the common factor.

$f (\frac{x + 3}{2 - 5 x}) = \frac{17 x}{- 5 x + 2} \cdot \frac{- 5 x + 2}{17}$

Step 5.3.6.3

Rewrite the expression.

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

Step 5.3.7

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

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

Step 5.3.8

Cancel the common factor of $- 5 x + 2$ .

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Step 5.3.8.1

Cancel the common factor.

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

Step 5.3.8.2

Divide $x$ by $1$ .

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

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x + 3}{2 - 5 x}$ is the inverse of $f (x) = \frac{2 x - 3}{5 x + 1}$ .

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