Find the Inverse (5x+15)/7

$\frac{5 x + 15}{7}$

Step 1

Interchange the variables.

$x = \frac{5 y + 15}{7}$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{5 y + 15}{7} = x$ .

$\frac{5 y + 15}{7} = x$

Multiply both sides by $7$ .

$\frac{5 y + 15}{7} \cdot 7 = x \cdot 7$

Simplify.

Simplify the left side.

Simplify $\frac{5 y + 15}{7} \cdot 7$ .

Factor $5$ out of $5 y + 15$ .

Factor $5$ out of $5 y$ .

$\frac{5 (y) + 15}{7} \cdot 7 = x \cdot 7$

Factor $5$ out of $15$ .

$\frac{5 y + 5 \cdot 3}{7} \cdot 7 = x \cdot 7$

Factor $5$ out of $5 y + 5 \cdot 3$ .

$\frac{5 (y + 3)}{7} \cdot 7 = x \cdot 7$

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{5 (y + 3)}{7} \cdot 7 = x \cdot 7$

Rewrite the expression.

$5 (y + 3) = x \cdot 7$

Apply the distributive property.

$5 y + 5 \cdot 3 = x \cdot 7$

Multiply $5$ by $3$ .

$5 y + 15 = x \cdot 7$

Simplify the right side.

Move $7$ to the left of $x$ .

$5 y + 15 = 7 x$

Solve for $y$ .

Subtract $15$ from both sides of the equation.

$5 y = 7 x - 15$

Divide each term in $5 y = 7 x - 15$ by $5$ and simplify.

Divide each term in $5 y = 7 x - 15$ by $5$ .

$\frac{5 y}{5} = \frac{7 x}{5} + \frac{- 15}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 y}{5} = \frac{7 x}{5} + \frac{- 15}{5}$

Divide $y$ by $1$ .

$y = \frac{7 x}{5} + \frac{- 15}{5}$

Simplify the right side.

Divide $- 15$ by $5$ .

$y = \frac{7 x}{5} - 3$

Step 3

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

$f^{- 1} (x) = \frac{7 x}{5} - 3$

Step 4

Verify if $f^{- 1} (x) = \frac{7 x}{5} - 3$ is the inverse of $f (x) = \frac{5 x + 15}{7}$ .

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 $f^{- 1} (\frac{5 x + 15}{7})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{5 x + 15}{7}) = \frac{7 (\frac{5 x + 15}{7})}{5} - 3$

Simplify each term.

Factor $5$ out of $5 x + 15$ .

Factor $5$ out of $5 x$ .

$f^{- 1} (\frac{5 x + 15}{7}) = \frac{7 (\frac{5 (x) + 15}{7})}{5} - 3$

Factor $5$ out of $15$ .

$f^{- 1} (\frac{5 x + 15}{7}) = \frac{7 (\frac{5 x + 5 \cdot 3}{7})}{5} - 3$

Factor $5$ out of $5 x + 5 \cdot 3$ .

$f^{- 1} (\frac{5 x + 15}{7}) = \frac{7 (\frac{5 (x + 3)}{7})}{5} - 3$

Combine $7$ and $\frac{5 (x + 3)}{7}$ .

$f^{- 1} (\frac{5 x + 15}{7}) = \frac{\frac{7 (5 (x + 3))}{7}}{5} - 3$

Multiply $7$ by $5$ .

$f^{- 1} (\frac{5 x + 15}{7}) = \frac{\frac{35 (x + 3)}{7}}{5} - 3$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{35 (x + 3)}{7}$ by cancelling the common factors.

Factor $7$ out of $35 (x + 3)$ .

$f^{- 1} (\frac{5 x + 15}{7}) = \frac{\frac{7 (5 (x + 3))}{7}}{5} - 3$

Factor $7$ out of $7$ .

$f^{- 1} (\frac{5 x + 15}{7}) = \frac{\frac{7 (5 (x + 3))}{7 (1)}}{5} - 3$

Cancel the common factor.

$f^{- 1} (\frac{5 x + 15}{7}) = \frac{\frac{7 (5 (x + 3))}{7 \cdot 1}}{5} - 3$

Rewrite the expression.

$f^{- 1} (\frac{5 x + 15}{7}) = \frac{\frac{5 (x + 3)}{1}}{5} - 3$

Divide $5 (x + 3)$ by $1$ .

$f^{- 1} (\frac{5 x + 15}{7}) = \frac{5 (x + 3)}{5} - 3$

Cancel the common factor of $5$ .

Cancel the common factor.

$f^{- 1} (\frac{5 x + 15}{7}) = \frac{5 (x + 3)}{5} - 3$

Divide $x + 3$ by $1$ .

$f^{- 1} (\frac{5 x + 15}{7}) = x + 3 - 3$

Combine the opposite terms in $x + 3 - 3$ .

Subtract $3$ from $3$ .

$f^{- 1} (\frac{5 x + 15}{7}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\frac{5 x + 15}{7}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{7 x}{5} - 3)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{7 x}{5} - 3) = \frac{5 (\frac{7 x}{5} - 3) + 15}{7}$

Simplify the numerator.

Factor $5$ out of $5 (\frac{7 x}{5} - 3) + 15$ .

Factor $5$ out of $15$ .

$f (\frac{7 x}{5} - 3) = \frac{5 (\frac{7 x}{5} - 3) + 5 \cdot 3}{7}$

Factor $5$ out of $5 (\frac{7 x}{5} - 3) + 5 \cdot 3$ .

$f (\frac{7 x}{5} - 3) = \frac{5 (\frac{7 x}{5} - 3 + 3)}{7}$

Add $- 3$ and $3$ .

$f (\frac{7 x}{5} - 3) = \frac{5 (\frac{7 x}{5} + 0)}{7}$

Add $\frac{7 x}{5}$ and $0$ .

$f (\frac{7 x}{5} - 3) = \frac{5 (\frac{7 x}{5})}{7}$

Combine $5$ and $\frac{7 x}{5}$ .

$f (\frac{7 x}{5} - 3) = \frac{\frac{5 (7 x)}{5}}{7}$

Multiply $5$ by $7$ .

$f (\frac{7 x}{5} - 3) = \frac{\frac{35 x}{5}}{7}$

Reduce the expression by cancelling the common factors.

Reduce the expression $\frac{35 x}{5}$ by cancelling the common factors.

Factor $5$ out of $35 x$ .

$f (\frac{7 x}{5} - 3) = \frac{\frac{5 (7 x)}{5}}{7}$

Factor $5$ out of $5$ .

$f (\frac{7 x}{5} - 3) = \frac{\frac{5 (7 x)}{5 (1)}}{7}$

Cancel the common factor.

$f (\frac{7 x}{5} - 3) = \frac{\frac{5 (7 x)}{5 \cdot 1}}{7}$

Rewrite the expression.

$f (\frac{7 x}{5} - 3) = \frac{\frac{7 x}{1}}{7}$

Divide $7 x$ by $1$ .

$f (\frac{7 x}{5} - 3) = \frac{7 x}{7}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f (\frac{7 x}{5} - 3) = \frac{7 x}{7}$

Divide $x$ by $1$ .

$f (\frac{7 x}{5} - 3) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{7 x}{5} - 3$ is the inverse of $f (x) = \frac{5 x + 15}{7}$ .

$f^{- 1} (x) = \frac{7 x}{5} - 3$

Find the Inverse (5x+15)/7

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