Find the Inverse y=(x+5)/7

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{y + 5}{7} = x$ .

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

Multiply both sides of the equation by $7$ .

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

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

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

Rewrite the expression.

$y + 5 = 7 x$

Subtract $5$ from both sides of the equation.

$y = 7 x - 5$

Step 3

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

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

Step 4

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

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

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

Cancel the common factor of $7$ .

Cancel the common factor.

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

Rewrite the expression.

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

Combine the opposite terms in $x + 5 - 5$ .

Subtract $5$ from $5$ .

$f^{- 1} (\frac{x + 5}{7}) = x + 0$

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Add $- 5$ and $5$ .

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

Add $7 x$ and $0$ .

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

Cancel the common factor of $7$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

$f (7 x - 5) = x$

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

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

Find the Inverse y=(x+5)/7

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