Find the Inverse A(b)=10*(b+7)/2

$A (b) = 10 \cdot \frac{b + 7}{2}$

Step 1

Write $A (b) = 10 \cdot \frac{b + 7}{2}$ as an equation.

$y = 10 \cdot \frac{b + 7}{2}$

Step 2

Interchange the variables.

$b = 10 \cdot \frac{y + 7}{2}$

Step 3

Solve for $y$ .

Rewrite the equation as $10 \cdot \frac{y + 7}{2} = b$ .

$10 \cdot \frac{y + 7}{2} = b$

Combine $10$ and $\frac{y + 7}{2}$ .

$\frac{10 (y + 7)}{2} = b$

Divide each term in $\frac{10 (y + 7)}{2} = b$ by $10$ and simplify.

Divide each term in $\frac{10 (y + 7)}{2} = b$ by $10$ .

$\frac{\frac{10 (y + 7)}{2}}{10} = \frac{b}{10}$

Simplify the left side.

Cancel the common factor of $10$ and $2$ .

Factor $2$ out of $10 (y + 7)$ .

$\frac{\frac{2 (5 (y + 7))}{2}}{10} = \frac{b}{10}$

Cancel the common factors.

Factor $2$ out of $2$ .

$\frac{\frac{2 (5 (y + 7))}{2 (1)}}{10} = \frac{b}{10}$

Cancel the common factor.

$\frac{\frac{2 (5 (y + 7))}{2 \cdot 1}}{10} = \frac{b}{10}$

Rewrite the expression.

$\frac{\frac{5 (y + 7)}{1}}{10} = \frac{b}{10}$

Divide $5 (y + 7)$ by $1$ .

$\frac{5 (y + 7)}{10} = \frac{b}{10}$

Cancel the common factors.

Factor $5$ out of $10$ .

$\frac{5 (y + 7)}{5 \cdot 2} = \frac{b}{10}$

Cancel the common factor.

$\frac{5 (y + 7)}{5 \cdot 2} = \frac{b}{10}$

Rewrite the expression.

$\frac{y + 7}{2} = \frac{b}{10}$

Multiply both sides of the equation by $2$ .

$2 \frac{y + 7}{2} = 2 \frac{b}{10}$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$2 \frac{y + 7}{2} = 2 \frac{b}{10}$

Rewrite the expression.

$y + 7 = 2 \frac{b}{10}$

Simplify the right side.

Cancel the common factor of $2$ .

Factor $2$ out of $10$ .

$y + 7 = 2 \frac{b}{2 (5)}$

Cancel the common factor.

$y + 7 = 2 \frac{b}{2 \cdot 5}$

Rewrite the expression.

$y + 7 = \frac{b}{5}$

Subtract $7$ from both sides of the equation.

$y = \frac{b}{5} - 7$

Step 4

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

$A^{- 1} (b) = \frac{b}{5} - 7$

Step 5

Verify if $A^{- 1} (b) = \frac{b}{5} - 7$ is the inverse of $A (b) = 10 \cdot \frac{b + 7}{2}$ .

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

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

Set up the composite result function.

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

Evaluate $A^{- 1} (10 \cdot \frac{b + 7}{2})$ by substituting in the value of $A$ into $A^{- 1}$ .

$A^{- 1} (10 \cdot \frac{b + 7}{2}) = \frac{10 \cdot \frac{b + 7}{2}}{5} - 7$

Simplify each term.

Cancel the common factor of $10$ and $5$ .

Factor $5$ out of $10 \cdot \frac{b + 7}{2}$ .

$A^{- 1} (10 \cdot \frac{b + 7}{2}) = \frac{5 (2 \cdot \frac{b + 7}{2})}{5} - 7$

Cancel the common factors.

Factor $5$ out of $5$ .

$A^{- 1} (10 \cdot \frac{b + 7}{2}) = \frac{5 (2 \cdot \frac{b + 7}{2})}{5 (1)} - 7$

Cancel the common factor.

$A^{- 1} (10 \cdot \frac{b + 7}{2}) = \frac{5 (2 \cdot \frac{b + 7}{2})}{5 \cdot 1} - 7$

Rewrite the expression.

$A^{- 1} (10 \cdot \frac{b + 7}{2}) = \frac{2 \cdot \frac{b + 7}{2}}{1} - 7$

Divide $2 \cdot \frac{b + 7}{2}$ by $1$ .

$A^{- 1} (10 \cdot \frac{b + 7}{2}) = 2 \cdot \frac{b + 7}{2} - 7$

Cancel the common factor of $2$ .

Cancel the common factor.

$A^{- 1} (10 \cdot \frac{b + 7}{2}) = 2 \cdot \frac{b + 7}{2} - 7$

Rewrite the expression.

$A^{- 1} (10 \cdot \frac{b + 7}{2}) = b + 7 - 7$

Combine the opposite terms in $b + 7 - 7$ .

Subtract $7$ from $7$ .

$A^{- 1} (10 \cdot \frac{b + 7}{2}) = b + 0$

Add $b$ and $0$ .

$A^{- 1} (10 \cdot \frac{b + 7}{2}) = b$

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

Set up the composite result function.

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

Evaluate $A (\frac{b}{5} - 7)$ by substituting in the value of $A^{- 1}$ into $A$ .

$A (\frac{b}{5} - 7) = 10 \cdot \frac{(\frac{b}{5} - 7) + 7}{2}$

Simplify the numerator.

Add $- 7$ and $7$ .

$A (\frac{b}{5} - 7) = 10 \cdot \frac{\frac{b}{5} + 0}{2}$

Add $\frac{b}{5}$ and $0$ .

$A (\frac{b}{5} - 7) = 10 \cdot \frac{\frac{b}{5}}{2}$

Simplify terms.

Cancel the common factor of $2$ .

Factor $2$ out of $10$ .

$A (\frac{b}{5} - 7) = 2 (5) \cdot \frac{\frac{b}{5}}{2}$

Cancel the common factor.

$A (\frac{b}{5} - 7) = 2 \cdot 5 \cdot \frac{\frac{b}{5}}{2}$

Rewrite the expression.

$A (\frac{b}{5} - 7) = 5 \cdot \frac{b}{5}$

$A (\frac{b}{5} - 7) = \frac{5 b}{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

$A (\frac{b}{5} - 7) = \frac{5 b}{5}$

Divide $b$ by $1$ .

$A (\frac{b}{5} - 7) = b$

Since $A^{- 1} (A (b)) = b$ and $A (A^{- 1} (b)) = b$ , then $A^{- 1} (b) = \frac{b}{5} - 7$ is the inverse of $A (b) = 10 \cdot \frac{b + 7}{2}$ .

$A^{- 1} (b) = \frac{b}{5} - 7$

Find the Inverse A(b)=10*(b+7)/2

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