Algebra Examples

$F (C) = \frac{9}{5} C + 32$

Step 1

Write $F (C) = \frac{9}{5} C + 32$ as an equation.

$y = \frac{9}{5} C + 32$

Step 2

Interchange the variables.

$C = \frac{9}{5} y + 32$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{9}{5} y + 32 = C$ .

$\frac{9}{5} y + 32 = C$

Step 3.2

Combine $\frac{9}{5}$ and $y$ .

$\frac{9 y}{5} + 32 = C$

Step 3.3

Subtract $32$ from both sides of the equation.

$\frac{9 y}{5} = C - 32$

Step 3.4

Multiply both sides of the equation by $\frac{5}{9}$ .

$\frac{5}{9} \cdot \frac{9 y}{5} = \frac{5}{9} (C - 32)$

Step 3.5

Simplify both sides of the equation.

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Step 3.5.1

Simplify the left side.

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Step 3.5.1.1

Simplify $\frac{5}{9} \cdot \frac{9 y}{5}$ .

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Step 3.5.1.1.1

Cancel the common factor of $5$ .

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Step 3.5.1.1.1.1

Cancel the common factor.

$\frac{5}{9} \cdot \frac{9 y}{5} = \frac{5}{9} (C - 32)$

Step 3.5.1.1.1.2

Rewrite the expression.

$\frac{1}{9} (9 y) = \frac{5}{9} (C - 32)$

Step 3.5.1.1.2

Cancel the common factor of $9$ .

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Step 3.5.1.1.2.1

Factor $9$ out of $9 y$ .

$\frac{1}{9} (9 (y)) = \frac{5}{9} (C - 32)$

Step 3.5.1.1.2.2

Cancel the common factor.

$\frac{1}{9} (9 y) = \frac{5}{9} (C - 32)$

Step 3.5.1.1.2.3

Rewrite the expression.

$y = \frac{5}{9} (C - 32)$

Step 3.5.2

Simplify the right side.

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Step 3.5.2.1

Simplify $\frac{5}{9} (C - 32)$ .

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Step 3.5.2.1.1

Apply the distributive property.

$y = \frac{5}{9} C + \frac{5}{9} \cdot - 32$

Step 3.5.2.1.2

Combine $\frac{5}{9}$ and $C$ .

$y = \frac{5 C}{9} + \frac{5}{9} \cdot - 32$

Step 3.5.2.1.3

Multiply $\frac{5}{9} \cdot - 32$ .

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Step 3.5.2.1.3.1

Combine $\frac{5}{9}$ and $- 32$ .

$y = \frac{5 C}{9} + \frac{5 \cdot - 32}{9}$

Step 3.5.2.1.3.2

Multiply $5$ by $- 32$ .

$y = \frac{5 C}{9} + \frac{- 160}{9}$

Step 3.5.2.1.4

Move the negative in front of the fraction.

$y = \frac{5 C}{9} - \frac{160}{9}$

Step 4

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

$F^{- 1} (C) = \frac{5 C}{9} - \frac{160}{9}$

Step 5

Verify if $F^{- 1} (C) = \frac{5 C}{9} - \frac{160}{9}$ is the inverse of $F (C) = \frac{9}{5} C + 32$ .

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Step 5.1

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

Step 5.2

Evaluate $F^{- 1} (F (C))$ .

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Step 5.2.1

Set up the composite result function.

$F^{- 1} (F (C))$

Step 5.2.2

Evaluate $F^{- 1} (\frac{9}{5} C + 32)$ by substituting in the value of $F$ into $F^{- 1}$ .

$F^{- 1} (\frac{9}{5} C + 32) = \frac{5 (\frac{9}{5} C + 32)}{9} - \frac{160}{9}$

Step 5.2.3

Combine the numerators over the common denominator.

$F^{- 1} (\frac{9}{5} C + 32) = \frac{5 (\frac{9}{5} C + 32) - 160}{9}$

Step 5.2.4

Simplify each term.

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Step 5.2.4.1

Combine $\frac{9}{5}$ and $C$ .

$F^{- 1} (\frac{9}{5} C + 32) = \frac{5 (\frac{9 C}{5} + 32) - 160}{9}$

Step 5.2.4.2

Apply the distributive property.

$F^{- 1} (\frac{9}{5} C + 32) = \frac{5 (\frac{9 C}{5}) + 5 \cdot 32 - 160}{9}$

Step 5.2.4.3

Cancel the common factor of $5$ .

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Step 5.2.4.3.1

Cancel the common factor.

$F^{- 1} (\frac{9}{5} C + 32) = \frac{5 (\frac{9 C}{5}) + 5 \cdot 32 - 160}{9}$

Step 5.2.4.3.2

Rewrite the expression.

$F^{- 1} (\frac{9}{5} C + 32) = \frac{9 C + 5 \cdot 32 - 160}{9}$

Step 5.2.4.4

Multiply $5$ by $32$ .

$F^{- 1} (\frac{9}{5} C + 32) = \frac{9 C + 160 - 160}{9}$

Step 5.2.5

Simplify terms.

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Step 5.2.5.1

Combine the opposite terms in $9 C + 160 - 160$ .

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Step 5.2.5.1.1

Subtract $160$ from $160$ .

$F^{- 1} (\frac{9}{5} C + 32) = \frac{9 C + 0}{9}$

Step 5.2.5.1.2

Add $9 C$ and $0$ .

$F^{- 1} (\frac{9}{5} C + 32) = \frac{9 C}{9}$

Step 5.2.5.2

Cancel the common factor of $9$ .

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Step 5.2.5.2.1

Cancel the common factor.

$F^{- 1} (\frac{9}{5} C + 32) = \frac{9 C}{9}$

Step 5.2.5.2.2

Divide $C$ by $1$ .

$F^{- 1} (\frac{9}{5} C + 32) = C$

Step 5.3

Evaluate $F (F^{- 1} (C))$ .

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Step 5.3.1

Set up the composite result function.

$F (F^{- 1} (C))$

Step 5.3.2

Evaluate $F (\frac{5 C}{9} - \frac{160}{9})$ by substituting in the value of $F^{- 1}$ into $F$ .

$F (\frac{5 C}{9} - \frac{160}{9}) = \frac{9}{5} \cdot (\frac{5 C}{9} - \frac{160}{9}) + 32$

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the distributive property.

$F (\frac{5 C}{9} - \frac{160}{9}) = \frac{9}{5} \cdot \frac{5 C}{9} + \frac{9}{5} \cdot (- \frac{160}{9}) + 32$

Step 5.3.3.2

Cancel the common factor of $9$ .

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Step 5.3.3.2.1

Cancel the common factor.

$F (\frac{5 C}{9} - \frac{160}{9}) = \frac{9}{5} \cdot \frac{5 C}{9} + \frac{9}{5} \cdot (- \frac{160}{9}) + 32$

Step 5.3.3.2.2

Rewrite the expression.

$F (\frac{5 C}{9} - \frac{160}{9}) = \frac{1}{5} \cdot (5 C) + \frac{9}{5} \cdot (- \frac{160}{9}) + 32$

Step 5.3.3.3

Cancel the common factor of $5$ .

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Step 5.3.3.3.1

Factor $5$ out of $5 C$ .

$F (\frac{5 C}{9} - \frac{160}{9}) = \frac{1}{5} \cdot (5 (C)) + \frac{9}{5} \cdot (- \frac{160}{9}) + 32$

Step 5.3.3.3.2

Cancel the common factor.

$F (\frac{5 C}{9} - \frac{160}{9}) = \frac{1}{5} \cdot (5 C) + \frac{9}{5} \cdot (- \frac{160}{9}) + 32$

Step 5.3.3.3.3

Rewrite the expression.

$F (\frac{5 C}{9} - \frac{160}{9}) = C + \frac{9}{5} \cdot (- \frac{160}{9}) + 32$

Step 5.3.3.4

Cancel the common factor of $9$ .

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Step 5.3.3.4.1

Move the leading negative in $- \frac{160}{9}$ into the numerator.

$F (\frac{5 C}{9} - \frac{160}{9}) = C + \frac{9}{5} \cdot \frac{- 160}{9} + 32$

Step 5.3.3.4.2

Cancel the common factor.

$F (\frac{5 C}{9} - \frac{160}{9}) = C + \frac{9}{5} \cdot \frac{- 160}{9} + 32$

Step 5.3.3.4.3

Rewrite the expression.

$F (\frac{5 C}{9} - \frac{160}{9}) = C + \frac{1}{5} \cdot - 160 + 32$

Step 5.3.3.5

Cancel the common factor of $5$ .

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Step 5.3.3.5.1

Factor $5$ out of $- 160$ .

$F (\frac{5 C}{9} - \frac{160}{9}) = C + \frac{1}{5} \cdot (5 (- 32)) + 32$

Step 5.3.3.5.2

Cancel the common factor.

$F (\frac{5 C}{9} - \frac{160}{9}) = C + \frac{1}{5} \cdot (5 \cdot - 32) + 32$

Step 5.3.3.5.3

Rewrite the expression.

$F (\frac{5 C}{9} - \frac{160}{9}) = C - 32 + 32$

Step 5.3.4

Combine the opposite terms in $C - 32 + 32$ .

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Step 5.3.4.1

Add $- 32$ and $32$ .

$F (\frac{5 C}{9} - \frac{160}{9}) = C + 0$

Step 5.3.4.2

Add $C$ and $0$ .

$F (\frac{5 C}{9} - \frac{160}{9}) = C$

Step 5.4

Since $F^{- 1} (F (C)) = C$ and $F (F^{- 1} (C)) = C$ , then $F^{- 1} (C) = \frac{5 C}{9} - \frac{160}{9}$ is the inverse of $F (C) = \frac{9}{5} C + 32$ .

$F^{- 1} (C) = \frac{5 C}{9} - \frac{160}{9}$