Find the Inverse a(n)=4n-25

$a (n) = 4 n - 25$

Step 1

Write $a (n) = 4 n - 25$ as an equation.

$y = 4 n - 25$

Step 2

Interchange the variables.

$n = 4 y - 25$

Step 3

Solve for $y$ .

Rewrite the equation as $4 y - 25 = n$ .

$4 y - 25 = n$

Add $25$ to both sides of the equation.

$4 y = n + 25$

Divide each term in $4 y = n + 25$ by $4$ and simplify.

Divide each term in $4 y = n + 25$ by $4$ .

$\frac{4 y}{4} = \frac{n}{4} + \frac{25}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{n}{4} + \frac{25}{4}$

Divide $y$ by $1$ .

$y = \frac{n}{4} + \frac{25}{4}$

Step 4

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

$a^{- 1} (n) = \frac{n}{4} + \frac{25}{4}$

Step 5

Verify if $a^{- 1} (n) = \frac{n}{4} + \frac{25}{4}$ is the inverse of $a (n) = 4 n - 25$ .

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 $a^{- 1} (4 n - 25)$ by substituting in the value of $a$ into $a^{- 1}$ .

$a^{- 1} (4 n - 25) = \frac{4 n - 25}{4} + \frac{25}{4}$

Combine the numerators over the common denominator.

$a^{- 1} (4 n - 25) = \frac{4 n - 25 + 25}{4}$

Combine the opposite terms in $4 n - 25 + 25$ .

Add $- 25$ and $25$ .

$a^{- 1} (4 n - 25) = \frac{4 n + 0}{4}$

Add $4 n$ and $0$ .

$a^{- 1} (4 n - 25) = \frac{4 n}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$a^{- 1} (4 n - 25) = \frac{4 n}{4}$

Divide $n$ by $1$ .

$a^{- 1} (4 n - 25) = n$

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

Set up the composite result function.

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

Evaluate $a (\frac{n}{4} + \frac{25}{4})$ by substituting in the value of $a^{- 1}$ into $a$ .

$a (\frac{n}{4} + \frac{25}{4}) = 4 (\frac{n}{4} + \frac{25}{4}) - 25$

Simplify each term.

Apply the distributive property.

$a (\frac{n}{4} + \frac{25}{4}) = 4 (\frac{n}{4}) + 4 (\frac{25}{4}) - 25$

Cancel the common factor of $4$ .

Cancel the common factor.

$a (\frac{n}{4} + \frac{25}{4}) = 4 (\frac{n}{4}) + 4 (\frac{25}{4}) - 25$

Rewrite the expression.

$a (\frac{n}{4} + \frac{25}{4}) = n + 4 (\frac{25}{4}) - 25$

Cancel the common factor of $4$ .

Cancel the common factor.

$a (\frac{n}{4} + \frac{25}{4}) = n + 4 (\frac{25}{4}) - 25$

Rewrite the expression.

$a (\frac{n}{4} + \frac{25}{4}) = n + 25 - 25$

Combine the opposite terms in $n + 25 - 25$ .

Subtract $25$ from $25$ .

$a (\frac{n}{4} + \frac{25}{4}) = n + 0$

Add $n$ and $0$ .

$a (\frac{n}{4} + \frac{25}{4}) = n$

Since $a^{- 1} (a (n)) = n$ and $a (a^{- 1} (n)) = n$ , then $a^{- 1} (n) = \frac{n}{4} + \frac{25}{4}$ is the inverse of $a (n) = 4 n - 25$ .

$a^{- 1} (n) = \frac{n}{4} + \frac{25}{4}$

Find the Inverse a(n)=4n-25

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