Find the Inverse c(n)=10n+20

$c (n) = 10 n + 20$

Step 1

Write $c (n) = 10 n + 20$ as an equation.

$y = 10 n + 20$

Step 2

Interchange the variables.

$n = 10 y + 20$

Step 3

Solve for $y$ .

Rewrite the equation as $10 y + 20 = n$ .

$10 y + 20 = n$

Subtract $20$ from both sides of the equation.

$10 y = n - 20$

Divide each term in $10 y = n - 20$ by $10$ and simplify.

Divide each term in $10 y = n - 20$ by $10$ .

$\frac{10 y}{10} = \frac{n}{10} + \frac{- 20}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 y}{10} = \frac{n}{10} + \frac{- 20}{10}$

Divide $y$ by $1$ .

$y = \frac{n}{10} + \frac{- 20}{10}$

Simplify the right side.

Divide $- 20$ by $10$ .

$y = \frac{n}{10} - 2$

Step 4

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

$c^{- 1} (n) = \frac{n}{10} - 2$

Step 5

Verify if $c^{- 1} (n) = \frac{n}{10} - 2$ is the inverse of $c (n) = 10 n + 20$ .

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

Evaluate $c^{- 1} (c (n))$ .

Set up the composite result function.

$c^{- 1} (c (n))$

Evaluate $c^{- 1} (10 n + 20)$ by substituting in the value of $c$ into $c^{- 1}$ .

$c^{- 1} (10 n + 20) = \frac{10 n + 20}{10} - 2$

Cancel the common factor of $10 n + 20$ and $10$ .

Factor $10$ out of $10 n$ .

$c^{- 1} (10 n + 20) = \frac{10 (n) + 20}{10} - 2$

Factor $10$ out of $20$ .

$c^{- 1} (10 n + 20) = \frac{10 (n) + 10 \cdot 2}{10} - 2$

Factor $10$ out of $10 (n) + 10 (2)$ .

$c^{- 1} (10 n + 20) = \frac{10 (n + 2)}{10} - 2$

Cancel the common factors.

Factor $10$ out of $10$ .

$c^{- 1} (10 n + 20) = \frac{10 (n + 2)}{10 (1)} - 2$

Cancel the common factor.

$c^{- 1} (10 n + 20) = \frac{10 (n + 2)}{10 \cdot 1} - 2$

Rewrite the expression.

$c^{- 1} (10 n + 20) = \frac{n + 2}{1} - 2$

Divide $n + 2$ by $1$ .

$c^{- 1} (10 n + 20) = n + 2 - 2$

Combine the opposite terms in $n + 2 - 2$ .

Subtract $2$ from $2$ .

$c^{- 1} (10 n + 20) = n + 0$

Add $n$ and $0$ .

$c^{- 1} (10 n + 20) = n$

Evaluate $c (c^{- 1} (n))$ .

Set up the composite result function.

$c (c^{- 1} (n))$

Evaluate $c (\frac{n}{10} - 2)$ by substituting in the value of $c^{- 1}$ into $c$ .

$c (\frac{n}{10} - 2) = 10 (\frac{n}{10} - 2) + 20$

Simplify each term.

Apply the distributive property.

$c (\frac{n}{10} - 2) = 10 (\frac{n}{10}) + 10 \cdot - 2 + 20$

Cancel the common factor of $10$ .

Cancel the common factor.

$c (\frac{n}{10} - 2) = 10 (\frac{n}{10}) + 10 \cdot - 2 + 20$

Rewrite the expression.

$c (\frac{n}{10} - 2) = n + 10 \cdot - 2 + 20$

Multiply $10$ by $- 2$ .

$c (\frac{n}{10} - 2) = n - 20 + 20$

Combine the opposite terms in $n - 20 + 20$ .

Add $- 20$ and $20$ .

$c (\frac{n}{10} - 2) = n + 0$

Add $n$ and $0$ .

$c (\frac{n}{10} - 2) = n$

Since $c^{- 1} (c (n)) = n$ and $c (c^{- 1} (n)) = n$ , then $c^{- 1} (n) = \frac{n}{10} - 2$ is the inverse of $c (n) = 10 n + 20$ .

$c^{- 1} (n) = \frac{n}{10} - 2$

Find the Inverse c(n)=10n+20

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