Find the Inverse a(n)=3n-20

$a (n) = 3 n - 20$

Step 1

Write $a (n) = 3 n - 20$ as an equation.

$y = 3 n - 20$

Step 2

Interchange the variables.

$n = 3 y - 20$

Step 3

Solve for $y$ .

Rewrite the equation as $3 y - 20 = n$ .

$3 y - 20 = n$

Add $20$ to both sides of the equation.

$3 y = n + 20$

Divide each term in $3 y = n + 20$ by $3$ and simplify.

Divide each term in $3 y = n + 20$ by $3$ .

$\frac{3 y}{3} = \frac{n}{3} + \frac{20}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{n}{3} + \frac{20}{3}$

Divide $y$ by $1$ .

$y = \frac{n}{3} + \frac{20}{3}$

Step 4

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

$a^{- 1} (n) = \frac{n}{3} + \frac{20}{3}$

Step 5

Verify if $a^{- 1} (n) = \frac{n}{3} + \frac{20}{3}$ is the inverse of $a (n) = 3 n - 20$ .

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 $a^{- 1} (3 n - 20)$ by substituting in the value of $a$ into $a^{- 1}$ .

$a^{- 1} (3 n - 20) = \frac{3 n - 20}{3} + \frac{20}{3}$

Combine the numerators over the common denominator.

$a^{- 1} (3 n - 20) = \frac{3 n - 20 + 20}{3}$

Combine the opposite terms in $3 n - 20 + 20$ .

Add $- 20$ and $20$ .

$a^{- 1} (3 n - 20) = \frac{3 n + 0}{3}$

Add $3 n$ and $0$ .

$a^{- 1} (3 n - 20) = \frac{3 n}{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$a^{- 1} (3 n - 20) = \frac{3 n}{3}$

Divide $n$ by $1$ .

$a^{- 1} (3 n - 20) = n$

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

Set up the composite result function.

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

Evaluate $a (\frac{n}{3} + \frac{20}{3})$ by substituting in the value of $a^{- 1}$ into $a$ .

$a (\frac{n}{3} + \frac{20}{3}) = 3 (\frac{n}{3} + \frac{20}{3}) - 20$

Simplify each term.

Apply the distributive property.

$a (\frac{n}{3} + \frac{20}{3}) = 3 (\frac{n}{3}) + 3 (\frac{20}{3}) - 20$

Cancel the common factor of $3$ .

Cancel the common factor.

$a (\frac{n}{3} + \frac{20}{3}) = 3 (\frac{n}{3}) + 3 (\frac{20}{3}) - 20$

Rewrite the expression.

$a (\frac{n}{3} + \frac{20}{3}) = n + 3 (\frac{20}{3}) - 20$

Cancel the common factor of $3$ .

Cancel the common factor.

$a (\frac{n}{3} + \frac{20}{3}) = n + 3 (\frac{20}{3}) - 20$

Rewrite the expression.

$a (\frac{n}{3} + \frac{20}{3}) = n + 20 - 20$

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

Subtract $20$ from $20$ .

$a (\frac{n}{3} + \frac{20}{3}) = n + 0$

Add $n$ and $0$ .

$a (\frac{n}{3} + \frac{20}{3}) = n$

Since $a^{- 1} (a (n)) = n$ and $a (a^{- 1} (n)) = n$ , then $a^{- 1} (n) = \frac{n}{3} + \frac{20}{3}$ is the inverse of $a (n) = 3 n - 20$ .

$a^{- 1} (n) = \frac{n}{3} + \frac{20}{3}$

Find the Inverse a(n)=3n-20

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