Algebra Examples

$f (x) = \frac{7 x + 2}{3} + 5$

Step 1

Write $f (x) = \frac{7 x + 2}{3} + 5$ as an equation.

$y = \frac{7 x + 2}{3} + 5$

Step 2

Interchange the variables.

$x = \frac{7 y + 2}{3} + 5$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{7 y + 2}{3} + 5 = x$ .

$\frac{7 y + 2}{3} + 5 = x$

Step 3.2

Simplify $\frac{7 y + 2}{3} + 5$ .

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Step 3.2.1

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

$\frac{7 y + 2}{3} + 5 \cdot \frac{3}{3} = x$

Step 3.2.2

Simplify terms.

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Step 3.2.2.1

Combine $5$ and $\frac{3}{3}$ .

$\frac{7 y + 2}{3} + \frac{5 \cdot 3}{3} = x$

Step 3.2.2.2

Combine the numerators over the common denominator.

$\frac{7 y + 2 + 5 \cdot 3}{3} = x$

Step 3.2.3

Simplify the numerator.

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Step 3.2.3.1

Multiply $5$ by $3$ .

$\frac{7 y + 2 + 15}{3} = x$

Step 3.2.3.2

Add $2$ and $15$ .

$\frac{7 y + 17}{3} = x$

Step 3.3

Multiply both sides by $3$ .

$\frac{7 y + 17}{3} \cdot 3 = x \cdot 3$

Step 3.4

Simplify.

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Step 3.4.1

Simplify the left side.

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Step 3.4.1.1

Cancel the common factor of $3$ .

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Step 3.4.1.1.1

Cancel the common factor.

$\frac{7 y + 17}{3} \cdot 3 = x \cdot 3$

Step 3.4.1.1.2

Rewrite the expression.

$7 y + 17 = x \cdot 3$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

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

$7 y + 17 = 3 x$

Step 3.5

Solve for $y$ .

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Step 3.5.1

Subtract $17$ from both sides of the equation.

$7 y = 3 x - 17$

Step 3.5.2

Divide each term in $7 y = 3 x - 17$ by $7$ and simplify.

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Step 3.5.2.1

Divide each term in $7 y = 3 x - 17$ by $7$ .

$\frac{7 y}{7} = \frac{3 x}{7} + \frac{- 17}{7}$

Step 3.5.2.2

Simplify the left side.

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Step 3.5.2.2.1

Cancel the common factor of $7$ .

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Step 3.5.2.2.1.1

Cancel the common factor.

$\frac{7 y}{7} = \frac{3 x}{7} + \frac{- 17}{7}$

Step 3.5.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{3 x}{7} + \frac{- 17}{7}$

Step 3.5.2.3

Simplify the right side.

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Step 3.5.2.3.1

Move the negative in front of the fraction.

$y = \frac{3 x}{7} - \frac{17}{7}$

Step 4

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

$f^{- 1} (x) = \frac{3 x}{7} - \frac{17}{7}$

Step 5

Verify if $f^{- 1} (x) = \frac{3 x}{7} - \frac{17}{7}$ is the inverse of $f (x) = \frac{7 x + 2}{3} + 5$ .

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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 $f^{- 1} (\frac{7 x + 2}{3} + 5)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 2}{3} + 5)}{7} - \frac{17}{7}$

Step 5.2.3

Combine the numerators over the common denominator.

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 2}{3} + 5) - 17}{7}$

Step 5.2.4

Simplify each term.

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Step 5.2.4.1

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

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 2}{3} + 5 \cdot \frac{3}{3}) - 17}{7}$

Step 5.2.4.2

Combine $5$ and $\frac{3}{3}$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 2}{3} + \frac{5 \cdot 3}{3}) - 17}{7}$

Step 5.2.4.3

Combine the numerators over the common denominator.

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 2 + 5 \cdot 3}{3}) - 17}{7}$

Step 5.2.4.4

Simplify the numerator.

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Step 5.2.4.4.1

Multiply $5$ by $3$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 2 + 15}{3}) - 17}{7}$

Step 5.2.4.4.2

Add $2$ and $15$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 17}{3}) - 17}{7}$

Step 5.2.4.5

Cancel the common factor of $3$ .

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Step 5.2.4.5.1

Cancel the common factor.

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 17}{3}) - 17}{7}$

Step 5.2.4.5.2

Rewrite the expression.

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{7 x + 17 - 17}{7}$

Step 5.2.5

Simplify terms.

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Step 5.2.5.1

Combine the opposite terms in $7 x + 17 - 17$ .

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Step 5.2.5.1.1

Subtract $17$ from $17$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{7 x + 0}{7}$

Step 5.2.5.1.2

Add $7 x$ and $0$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{7 x}{7}$

Step 5.2.5.2

Cancel the common factor of $7$ .

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Step 5.2.5.2.1

Cancel the common factor.

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{7 x}{7}$

Step 5.2.5.2.2

Divide $x$ by $1$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = x$

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 $f (\frac{3 x}{7} - \frac{17}{7})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{7 (\frac{3 x}{7} - \frac{17}{7}) + 2}{3} + 5$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Simplify the numerator.

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Step 5.3.3.1.1

Apply the distributive property.

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{7 (\frac{3 x}{7}) + 7 (- \frac{17}{7}) + 2}{3} + 5$

Step 5.3.3.1.2

Cancel the common factor of $7$ .

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Step 5.3.3.1.2.1

Cancel the common factor.

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{7 (\frac{3 x}{7}) + 7 (- \frac{17}{7}) + 2}{3} + 5$

Step 5.3.3.1.2.2

Rewrite the expression.

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 x + 7 (- \frac{17}{7}) + 2}{3} + 5$

Step 5.3.3.1.3

Cancel the common factor of $7$ .

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Step 5.3.3.1.3.1

Move the leading negative in $- \frac{17}{7}$ into the numerator.

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 x + 7 (\frac{- 17}{7}) + 2}{3} + 5$

Step 5.3.3.1.3.2

Cancel the common factor.

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 x + 7 (\frac{- 17}{7}) + 2}{3} + 5$

Step 5.3.3.1.3.3

Rewrite the expression.

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 x - 17 + 2}{3} + 5$

Step 5.3.3.1.4

Add $- 17$ and $2$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 x - 15}{3} + 5$

Step 5.3.3.1.5

Factor $3$ out of $3 x - 15$ .

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Step 5.3.3.1.5.1

Factor $3$ out of $3 x$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 (x) - 15}{3} + 5$

Step 5.3.3.1.5.2

Factor $3$ out of $- 15$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 x + 3 \cdot - 5}{3} + 5$

Step 5.3.3.1.5.3

Factor $3$ out of $3 x + 3 \cdot - 5$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 (x - 5)}{3} + 5$

Step 5.3.3.2

Cancel the common factor of $3$ .

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Step 5.3.3.2.1

Cancel the common factor.

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 (x - 5)}{3} + 5$

Step 5.3.3.2.2

Divide $x - 5$ by $1$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = x - 5 + 5$

Step 5.3.4

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

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Step 5.3.4.1

Add $- 5$ and $5$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = x + 0$

Step 5.3.4.2

Add $x$ and $0$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{3 x}{7} - \frac{17}{7}$ is the inverse of $f (x) = \frac{7 x + 2}{3} + 5$ .

$f^{- 1} (x) = \frac{3 x}{7} - \frac{17}{7}$