Find the Inverse 10*(x+7)/2

$10 \cdot \frac{x + 7}{2}$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

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

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

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

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

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

Simplify the left side.

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

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

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Cancel the common factors.

Factor $5$ out of $10$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply both sides of the equation by $2$ .

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

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Cancel the common factor of $2$ .

Factor $2$ out of $10$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Subtract $7$ from both sides of the equation.

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

Step 3

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

$f^{- 1} (x) = \frac{x}{5} - 7$

Step 4

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

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

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

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

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

Simplify each term.

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

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

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

Cancel the common factors.

Factor $5$ out of $5$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Subtract $7$ from $7$ .

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

Add $x$ and $0$ .

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

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

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

$f (\frac{x}{5} - 7) = 10 \cdot \frac{(\frac{x}{5} - 7) + 7}{2}$

Simplify the numerator.

Add $- 7$ and $7$ .

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

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

$f (\frac{x}{5} - 7) = 10 \cdot \frac{\frac{x}{5}}{2}$

Simplify terms.

Cancel the common factor of $2$ .

Factor $2$ out of $10$ .

$f (\frac{x}{5} - 7) = 2 (5) \cdot \frac{\frac{x}{5}}{2}$

Cancel the common factor.

$f (\frac{x}{5} - 7) = 2 \cdot 5 \cdot \frac{\frac{x}{5}}{2}$

Rewrite the expression.

$f (\frac{x}{5} - 7) = 5 \cdot \frac{x}{5}$

$f (\frac{x}{5} - 7) = \frac{5 x}{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (\frac{x}{5} - 7) = \frac{5 x}{5}$

Divide $x$ by $1$ .

$f (\frac{x}{5} - 7) = x$

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

$f^{- 1} (x) = \frac{x}{5} - 7$

Find the Inverse 10*(x+7)/2

Related Questions

Support LCMGCF.com by disabling your adblocker.