Find the Inverse A(n)=3n-15

$A (n) = 3 n - 15$

Step 1

Write $A (n) = 3 n - 15$ as an equation.

$y = 3 n - 15$

Step 2

Interchange the variables.

$n = 3 y - 15$

Step 3

Solve for $y$ .

Rewrite the equation as $3 y - 15 = n$ .

$3 y - 15 = n$

Add $15$ to both sides of the equation.

$3 y = n + 15$

Divide each term in $3 y = n + 15$ by $3$ and simplify.

Divide each term in $3 y = n + 15$ by $3$ .

$\frac{3 y}{3} = \frac{n}{3} + \frac{15}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{n}{3} + \frac{15}{3}$

Divide $y$ by $1$ .

$y = \frac{n}{3} + \frac{15}{3}$

Simplify the right side.

Divide $15$ by $3$ .

$y = \frac{n}{3} + 5$

Step 4

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

$A^{- 1} (n) = \frac{n}{3} + 5$

Step 5

Verify if $A^{- 1} (n) = \frac{n}{3} + 5$ is the inverse of $A (n) = 3 n - 15$ .

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

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

Set up the composite result function.

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

Evaluate $A^{- 1} (3 n - 15)$ by substituting in the value of $A$ into $A^{- 1}$ .

$A^{- 1} (3 n - 15) = \frac{3 n - 15}{3} + 5$

Cancel the common factor of $3 n - 15$ and $3$ .

Factor $3$ out of $3 n$ .

$A^{- 1} (3 n - 15) = \frac{3 (n) - 15}{3} + 5$

Factor $3$ out of $- 15$ .

$A^{- 1} (3 n - 15) = \frac{3 (n) + 3 \cdot - 5}{3} + 5$

Factor $3$ out of $3 (n) + 3 (- 5)$ .

$A^{- 1} (3 n - 15) = \frac{3 (n - 5)}{3} + 5$

Cancel the common factors.

Factor $3$ out of $3$ .

$A^{- 1} (3 n - 15) = \frac{3 (n - 5)}{3 (1)} + 5$

Cancel the common factor.

$A^{- 1} (3 n - 15) = \frac{3 (n - 5)}{3 \cdot 1} + 5$

Rewrite the expression.

$A^{- 1} (3 n - 15) = \frac{n - 5}{1} + 5$

Divide $n - 5$ by $1$ .

$A^{- 1} (3 n - 15) = n - 5 + 5$

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

Add $- 5$ and $5$ .

$A^{- 1} (3 n - 15) = n + 0$

Add $n$ and $0$ .

$A^{- 1} (3 n - 15) = n$

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

Set up the composite result function.

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

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

$A (\frac{n}{3} + 5) = 3 (\frac{n}{3} + 5) - 15$

Simplify each term.

Apply the distributive property.

$A (\frac{n}{3} + 5) = 3 (\frac{n}{3}) + 3 \cdot 5 - 15$

Cancel the common factor of $3$ .

Cancel the common factor.

$A (\frac{n}{3} + 5) = 3 (\frac{n}{3}) + 3 \cdot 5 - 15$

Rewrite the expression.

$A (\frac{n}{3} + 5) = n + 3 \cdot 5 - 15$

Multiply $3$ by $5$ .

$A (\frac{n}{3} + 5) = n + 15 - 15$

Combine the opposite terms in $n + 15 - 15$ .

Subtract $15$ from $15$ .

$A (\frac{n}{3} + 5) = n + 0$

Add $n$ and $0$ .

$A (\frac{n}{3} + 5) = n$

Since $A^{- 1} (A (n)) = n$ and $A (A^{- 1} (n)) = n$ , then $A^{- 1} (n) = \frac{n}{3} + 5$ is the inverse of $A (n) = 3 n - 15$ .

$A^{- 1} (n) = \frac{n}{3} + 5$

Find the Inverse A(n)=3n-15

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