Find the Inverse C(n)=20n+15

$C (n) = 20 n + 15$

Step 1

Write $C (n) = 20 n + 15$ as an equation.

$y = 20 n + 15$

Step 2

Interchange the variables.

$n = 20 y + 15$

Step 3

Solve for $y$ .

Rewrite the equation as $20 y + 15 = n$ .

$20 y + 15 = n$

Subtract $15$ from both sides of the equation.

$20 y = n - 15$

Divide each term in $20 y = n - 15$ by $20$ and simplify.

Divide each term in $20 y = n - 15$ by $20$ .

$\frac{20 y}{20} = \frac{n}{20} + \frac{- 15}{20}$

Simplify the left side.

Cancel the common factor of $20$ .

Cancel the common factor.

$\frac{20 y}{20} = \frac{n}{20} + \frac{- 15}{20}$

Divide $y$ by $1$ .

$y = \frac{n}{20} + \frac{- 15}{20}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 15$ and $20$ .

Factor $5$ out of $- 15$ .

$y = \frac{n}{20} + \frac{5 (- 3)}{20}$

Cancel the common factors.

Factor $5$ out of $20$ .

$y = \frac{n}{20} + \frac{5 \cdot - 3}{5 \cdot 4}$

Cancel the common factor.

$y = \frac{n}{20} + \frac{5 \cdot - 3}{5 \cdot 4}$

Rewrite the expression.

$y = \frac{n}{20} + \frac{- 3}{4}$

Move the negative in front of the fraction.

$y = \frac{n}{20} - \frac{3}{4}$

Step 4

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

$C^{- 1} (n) = \frac{n}{20} - \frac{3}{4}$

Step 5

Verify if $C^{- 1} (n) = \frac{n}{20} - \frac{3}{4}$ is the inverse of $C (n) = 20 n + 15$ .

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 $C^{- 1} (20 n + 15)$ by substituting in the value of $C$ into $C^{- 1}$ .

$C^{- 1} (20 n + 15) = \frac{20 n + 15}{20} - \frac{3}{4}$

Cancel the common factor of $20 n + 15$ and $20$ .

Factor $5$ out of $20 n$ .

$C^{- 1} (20 n + 15) = \frac{5 (4 n) + 15}{20} - \frac{3}{4}$

Factor $5$ out of $15$ .

$C^{- 1} (20 n + 15) = \frac{5 (4 n) + 5 (3)}{20} - \frac{3}{4}$

Factor $5$ out of $5 (4 n) + 5 (3)$ .

$C^{- 1} (20 n + 15) = \frac{5 (4 n + 3)}{20} - \frac{3}{4}$

Cancel the common factors.

Factor $5$ out of $20$ .

$C^{- 1} (20 n + 15) = \frac{5 (4 n + 3)}{5 (4)} - \frac{3}{4}$

Cancel the common factor.

$C^{- 1} (20 n + 15) = \frac{5 (4 n + 3)}{5 \cdot 4} - \frac{3}{4}$

Rewrite the expression.

$C^{- 1} (20 n + 15) = \frac{4 n + 3}{4} - \frac{3}{4}$

Combine the numerators over the common denominator.

$C^{- 1} (20 n + 15) = \frac{4 n + 3 - 3}{4}$

Combine the opposite terms in $4 n + 3 - 3$ .

Subtract $3$ from $3$ .

$C^{- 1} (20 n + 15) = \frac{4 n + 0}{4}$

Add $4 n$ and $0$ .

$C^{- 1} (20 n + 15) = \frac{4 n}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$C^{- 1} (20 n + 15) = \frac{4 n}{4}$

Divide $n$ by $1$ .

$C^{- 1} (20 n + 15) = n$

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

Set up the composite result function.

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

Evaluate $C (\frac{n}{20} - \frac{3}{4})$ by substituting in the value of $C^{- 1}$ into $C$ .

$C (\frac{n}{20} - \frac{3}{4}) = 20 (\frac{n}{20} - \frac{3}{4}) + 15$

Simplify each term.

Apply the distributive property.

$C (\frac{n}{20} - \frac{3}{4}) = 20 (\frac{n}{20}) + 20 (- \frac{3}{4}) + 15$

Cancel the common factor of $20$ .

Cancel the common factor.

$C (\frac{n}{20} - \frac{3}{4}) = 20 (\frac{n}{20}) + 20 (- \frac{3}{4}) + 15$

Rewrite the expression.

$C (\frac{n}{20} - \frac{3}{4}) = n + 20 (- \frac{3}{4}) + 15$

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{3}{4}$ into the numerator.

$C (\frac{n}{20} - \frac{3}{4}) = n + 20 (\frac{- 3}{4}) + 15$

Factor $4$ out of $20$ .

$C (\frac{n}{20} - \frac{3}{4}) = n + 4 (5) (\frac{- 3}{4}) + 15$

Cancel the common factor.

$C (\frac{n}{20} - \frac{3}{4}) = n + 4 \cdot (5 (\frac{- 3}{4})) + 15$

Rewrite the expression.

$C (\frac{n}{20} - \frac{3}{4}) = n + 5 \cdot - 3 + 15$

Multiply $5$ by $- 3$ .

$C (\frac{n}{20} - \frac{3}{4}) = n - 15 + 15$

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

Add $- 15$ and $15$ .

$C (\frac{n}{20} - \frac{3}{4}) = n + 0$

Add $n$ and $0$ .

$C (\frac{n}{20} - \frac{3}{4}) = n$

Since $C^{- 1} (C (n)) = n$ and $C (C^{- 1} (n)) = n$ , then $C^{- 1} (n) = \frac{n}{20} - \frac{3}{4}$ is the inverse of $C (n) = 20 n + 15$ .

$C^{- 1} (n) = \frac{n}{20} - \frac{3}{4}$

Find the Inverse C(n)=20n+15

Related Questions

Support LCMGCF.com by disabling your adblocker.