Finite Math Examples

$y = \frac{7 x}{5} + \frac{16}{5}$

Step 1

Interchange the variables.

$x = \frac{7 y}{5} + \frac{16}{5}$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\frac{7 y}{5} + \frac{16}{5} = x$ .

$\frac{7 y}{5} + \frac{16}{5} = x$

Step 2.2

Subtract $\frac{16}{5}$ from both sides of the equation.

$\frac{7 y}{5} = x - \frac{16}{5}$

Step 2.3

Multiply both sides of the equation by $\frac{5}{7}$ .

$\frac{5}{7} \cdot \frac{7 y}{5} = \frac{5}{7} (x - \frac{16}{5})$

Step 2.4

Simplify both sides of the equation.

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Step 2.4.1

Simplify the left side.

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Step 2.4.1.1

Simplify $\frac{5}{7} \cdot \frac{7 y}{5}$ .

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Step 2.4.1.1.1

Cancel the common factor of $5$ .

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Step 2.4.1.1.1.1

Cancel the common factor.

$\frac{5}{7} \cdot \frac{7 y}{5} = \frac{5}{7} (x - \frac{16}{5})$

Step 2.4.1.1.1.2

Rewrite the expression.

$\frac{1}{7} (7 y) = \frac{5}{7} (x - \frac{16}{5})$

Step 2.4.1.1.2

Cancel the common factor of $7$ .

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Step 2.4.1.1.2.1

Factor $7$ out of $7 y$ .

$\frac{1}{7} (7 (y)) = \frac{5}{7} (x - \frac{16}{5})$

Step 2.4.1.1.2.2

Cancel the common factor.

$\frac{1}{7} (7 y) = \frac{5}{7} (x - \frac{16}{5})$

Step 2.4.1.1.2.3

Rewrite the expression.

$y = \frac{5}{7} (x - \frac{16}{5})$

Step 2.4.2

Simplify the right side.

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Step 2.4.2.1

Simplify $\frac{5}{7} (x - \frac{16}{5})$ .

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Step 2.4.2.1.1

Apply the distributive property.

$y = \frac{5}{7} x + \frac{5}{7} (- \frac{16}{5})$

Step 2.4.2.1.2

Combine $\frac{5}{7}$ and $x$ .

$y = \frac{5 x}{7} + \frac{5}{7} (- \frac{16}{5})$

Step 2.4.2.1.3

Cancel the common factor of $5$ .

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Step 2.4.2.1.3.1

Move the leading negative in $- \frac{16}{5}$ into the numerator.

$y = \frac{5 x}{7} + \frac{5}{7} \cdot \frac{- 16}{5}$

Step 2.4.2.1.3.2

Cancel the common factor.

$y = \frac{5 x}{7} + \frac{5}{7} \cdot \frac{- 16}{5}$

Step 2.4.2.1.3.3

Rewrite the expression.

$y = \frac{5 x}{7} + \frac{1}{7} \cdot - 16$

Step 2.4.2.1.4

Combine $\frac{1}{7}$ and $- 16$ .

$y = \frac{5 x}{7} + \frac{- 16}{7}$

Step 2.4.2.1.5

Move the negative in front of the fraction.

$y = \frac{5 x}{7} - \frac{16}{7}$

Step 3

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

$f^{- 1} (x) = \frac{5 x}{7} - \frac{16}{7}$

Step 4

Verify if $f^{- 1} (x) = \frac{5 x}{7} - \frac{16}{7}$ is the inverse of $f (x) = \frac{7 x}{5} + \frac{16}{5}$ .

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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 $f^{- 1} (\frac{7 x}{5} + \frac{16}{5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{7 x}{5} + \frac{16}{5}) = \frac{5 (\frac{7 x}{5} + \frac{16}{5})}{7} - \frac{16}{7}$

Step 4.2.3

Combine the numerators over the common denominator.

$f^{- 1} (\frac{7 x}{5} + \frac{16}{5}) = \frac{5 (\frac{7 x}{5} + \frac{16}{5}) - 16}{7}$

Step 4.2.4

Simplify each term.

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Step 4.2.4.1

Apply the distributive property.

$f^{- 1} (\frac{7 x}{5} + \frac{16}{5}) = \frac{5 (\frac{7 x}{5}) + 5 (\frac{16}{5}) - 16}{7}$

Step 4.2.4.2

Cancel the common factor of $5$ .

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Step 4.2.4.2.1

Cancel the common factor.

$f^{- 1} (\frac{7 x}{5} + \frac{16}{5}) = \frac{5 (\frac{7 x}{5}) + 5 (\frac{16}{5}) - 16}{7}$

Step 4.2.4.2.2

Rewrite the expression.

$f^{- 1} (\frac{7 x}{5} + \frac{16}{5}) = \frac{7 x + 5 (\frac{16}{5}) - 16}{7}$

Step 4.2.4.3

Cancel the common factor of $5$ .

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Step 4.2.4.3.1

Cancel the common factor.

$f^{- 1} (\frac{7 x}{5} + \frac{16}{5}) = \frac{7 x + 5 (\frac{16}{5}) - 16}{7}$

Step 4.2.4.3.2

Rewrite the expression.

$f^{- 1} (\frac{7 x}{5} + \frac{16}{5}) = \frac{7 x + 16 - 16}{7}$

Step 4.2.5

Simplify terms.

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Step 4.2.5.1

Combine the opposite terms in $7 x + 16 - 16$ .

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Step 4.2.5.1.1

Subtract $16$ from $16$ .

$f^{- 1} (\frac{7 x}{5} + \frac{16}{5}) = \frac{7 x + 0}{7}$

Step 4.2.5.1.2

Add $7 x$ and $0$ .

$f^{- 1} (\frac{7 x}{5} + \frac{16}{5}) = \frac{7 x}{7}$

Step 4.2.5.2

Cancel the common factor of $7$ .

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Step 4.2.5.2.1

Cancel the common factor.

$f^{- 1} (\frac{7 x}{5} + \frac{16}{5}) = \frac{7 x}{7}$

Step 4.2.5.2.2

Divide $x$ by $1$ .

$f^{- 1} (\frac{7 x}{5} + \frac{16}{5}) = x$

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 $f (\frac{5 x}{7} - \frac{16}{7})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{5 x}{7} - \frac{16}{7}) = \frac{7 (\frac{5 x}{7} - \frac{16}{7})}{5} + \frac{16}{5}$

Step 4.3.3

Combine the numerators over the common denominator.

$f (\frac{5 x}{7} - \frac{16}{7}) = \frac{7 (\frac{5 x}{7} - \frac{16}{7}) + 16}{5}$

Step 4.3.4

Simplify each term.

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Step 4.3.4.1

Apply the distributive property.

$f (\frac{5 x}{7} - \frac{16}{7}) = \frac{7 (\frac{5 x}{7}) + 7 (- \frac{16}{7}) + 16}{5}$

Step 4.3.4.2

Cancel the common factor of $7$ .

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Step 4.3.4.2.1

Cancel the common factor.

$f (\frac{5 x}{7} - \frac{16}{7}) = \frac{7 (\frac{5 x}{7}) + 7 (- \frac{16}{7}) + 16}{5}$

Step 4.3.4.2.2

Rewrite the expression.

$f (\frac{5 x}{7} - \frac{16}{7}) = \frac{5 x + 7 (- \frac{16}{7}) + 16}{5}$

Step 4.3.4.3

Cancel the common factor of $7$ .

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Step 4.3.4.3.1

Move the leading negative in $- \frac{16}{7}$ into the numerator.

$f (\frac{5 x}{7} - \frac{16}{7}) = \frac{5 x + 7 (\frac{- 16}{7}) + 16}{5}$

Step 4.3.4.3.2

Cancel the common factor.

$f (\frac{5 x}{7} - \frac{16}{7}) = \frac{5 x + 7 (\frac{- 16}{7}) + 16}{5}$

Step 4.3.4.3.3

Rewrite the expression.

$f (\frac{5 x}{7} - \frac{16}{7}) = \frac{5 x - 16 + 16}{5}$

Step 4.3.5

Simplify terms.

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Step 4.3.5.1

Combine the opposite terms in $5 x - 16 + 16$ .

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Step 4.3.5.1.1

Add $- 16$ and $16$ .

$f (\frac{5 x}{7} - \frac{16}{7}) = \frac{5 x + 0}{5}$

Step 4.3.5.1.2

Add $5 x$ and $0$ .

$f (\frac{5 x}{7} - \frac{16}{7}) = \frac{5 x}{5}$

Step 4.3.5.2

Cancel the common factor of $5$ .

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Step 4.3.5.2.1

Cancel the common factor.

$f (\frac{5 x}{7} - \frac{16}{7}) = \frac{5 x}{5}$

Step 4.3.5.2.2

Divide $x$ by $1$ .

$f (\frac{5 x}{7} - \frac{16}{7}) = x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{5 x}{7} - \frac{16}{7}$ is the inverse of $f (x) = \frac{7 x}{5} + \frac{16}{5}$ .

$f^{- 1} (x) = \frac{5 x}{7} - \frac{16}{7}$