Algebra Examples

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

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 $y$ .

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Step 3.1

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

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

Step 3.2

Simplify $10 \cdot \frac{y + 7}{2}$ .

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Step 3.2.1

Cancel the common factor of $2$ .

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Step 3.2.1.1

Factor $2$ out of $10$ .

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

Step 3.2.1.2

Cancel the common factor.

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

Step 3.2.1.3

Rewrite the expression.

$5 \cdot (y + 7) = b$

Step 3.2.2

Apply the distributive property.

$5 y + 5 \cdot 7 = b$

Step 3.2.3

Multiply $5$ by $7$ .

$5 y + 35 = b$

Step 3.3

Subtract $35$ from both sides of the equation.

$5 y = b - 35$

Step 3.4

Divide each term in $5 y = b - 35$ by $5$ and simplify.

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Step 3.4.1

Divide each term in $5 y = b - 35$ by $5$ .

$\frac{5 y}{5} = \frac{b}{5} + \frac{- 35}{5}$

Step 3.4.2

Simplify the left side.

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Step 3.4.2.1

Cancel the common factor of $5$ .

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Step 3.4.2.1.1

Cancel the common factor.

$\frac{5 y}{5} = \frac{b}{5} + \frac{- 35}{5}$

Step 3.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{b}{5} + \frac{- 35}{5}$

Step 3.4.3

Simplify the right side.

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Step 3.4.3.1

Divide $- 35$ by $5$ .

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

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

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Step 5.1

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

Step 5.2

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

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

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

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

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

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

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Step 5.2.3.1.1

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

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

Step 5.2.3.1.2

Cancel the common factors.

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Step 5.2.3.1.2.1

Factor $5$ out of $5$ .

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

Step 5.2.3.1.2.2

Cancel the common factor.

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

Step 5.2.3.1.2.3

Rewrite the expression.

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

Step 5.2.3.1.2.4

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

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

Step 5.2.3.2

Cancel the common factor of $2$ .

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Step 5.2.3.2.1

Cancel the common factor.

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

Step 5.2.3.2.2

Rewrite the expression.

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

Step 5.2.4

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

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Step 5.2.4.1

Subtract $7$ from $7$ .

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

Step 5.2.4.2

Add $b$ and $0$ .

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

Step 5.3

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

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

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

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

Step 5.3.3

Simplify the numerator.

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Step 5.3.3.1

Add $- 7$ and $7$ .

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

Step 5.3.3.2

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

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

Step 5.3.4

Simplify terms.

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Step 5.3.4.1

Cancel the common factor of $2$ .

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Step 5.3.4.1.1

Factor $2$ out of $10$ .

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

Step 5.3.4.1.2

Cancel the common factor.

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

Step 5.3.4.1.3

Rewrite the expression.

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

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

Step 5.3.4.2

Cancel the common factor of $5$ .

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Step 5.3.4.2.1

Cancel the common factor.

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

Step 5.3.4.2.2

Divide $b$ by $1$ .

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

Step 5.4

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

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