Find the Inverse f(x)=(7x+2)/3+5

$f (x) = \frac{7 x + 2}{3} + 5$

Write $f (x) = \frac{7 x + 2}{3} + 5$ as an equation.

$y = \frac{7 x + 2}{3} + 5$

Interchange the variables.

$x = \frac{7 y + 2}{3} + 5$

Solve for $y$ .

Rewrite the equation as $\frac{7 y + 2}{3} + 5 = x$ .

$\frac{7 y + 2}{3} + 5 = x$

Simplify $\frac{7 y + 2}{3} + 5$ .

To write $5$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{7 y + 2}{3} + 5 \cdot \frac{3}{3} = x$

Simplify terms.

Combine $5$ and $\frac{3}{3}$ .

$\frac{7 y + 2}{3} + \frac{5 \cdot 3}{3} = x$

Combine the numerators over the common denominator.

$\frac{7 y + 2 + 5 \cdot 3}{3} = x$

Simplify the numerator.

Multiply $5$ by $3$ .

$\frac{7 y + 2 + 15}{3} = x$

Add $2$ and $15$ .

$\frac{7 y + 17}{3} = x$

Multiply both sides by $3$ .

$\frac{7 y + 17}{3} \cdot 3 = x \cdot 3$

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{7 y + 17}{3} \cdot 3 = x \cdot 3$

Rewrite the expression.

$7 y + 17 = x \cdot 3$

Simplify the right side.

Move $3$ to the left of $x$ .

$7 y + 17 = 3 x$

Solve for $y$ .

Subtract $17$ from both sides of the equation.

$7 y = 3 x - 17$

Divide each term in $7 y = 3 x - 17$ by $7$ and simplify.

Divide each term in $7 y = 3 x - 17$ by $7$ .

$\frac{7 y}{7} = \frac{3 x}{7} + \frac{- 17}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 y}{7} = \frac{3 x}{7} + \frac{- 17}{7}$

Divide $y$ by $1$ .

$y = \frac{3 x}{7} + \frac{- 17}{7}$

Simplify the right side.

Move the negative in front of the fraction.

$y = \frac{3 x}{7} - \frac{17}{7}$

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

$f^{- 1} (x) = \frac{3 x}{7} - \frac{17}{7}$

Verify if $f^{- 1} (x) = \frac{3 x}{7} - \frac{17}{7}$ is the inverse of $f (x) = \frac{7 x + 2}{3} + 5$ .

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 $f^{- 1} (\frac{7 x + 2}{3} + 5)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 2}{3} + 5)}{7} - \frac{17}{7}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 2}{3} + 5) - 17}{7}$

Simplify each term.

To write $5$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 2}{3} + 5 \cdot \frac{3}{3}) - 17}{7}$

Combine $5$ and $\frac{3}{3}$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 2}{3} + \frac{5 \cdot 3}{3}) - 17}{7}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 2 + 5 \cdot 3}{3}) - 17}{7}$

Simplify the numerator.

Multiply $5$ by $3$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 2 + 15}{3}) - 17}{7}$

Add $2$ and $15$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 17}{3}) - 17}{7}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{3 (\frac{7 x + 17}{3}) - 17}{7}$

Rewrite the expression.

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{7 x + 17 - 17}{7}$

Simplify terms.

Combine the opposite terms in $7 x + 17 - 17$ .

Subtract $17$ from $17$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{7 x + 0}{7}$

Add $7 x$ and $0$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{7 x}{7}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f^{- 1} (\frac{7 x + 2}{3} + 5) = \frac{7 x}{7}$

Divide $x$ by $1$ .

$f^{- 1} (\frac{7 x + 2}{3} + 5) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{3 x}{7} - \frac{17}{7})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{7 (\frac{3 x}{7} - \frac{17}{7}) + 2}{3} + 5$

Simplify each term.

Simplify the numerator.

Apply the distributive property.

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{7 (\frac{3 x}{7}) + 7 (- \frac{17}{7}) + 2}{3} + 5$

Cancel the common factor of $7$ .

Cancel the common factor.

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{7 (\frac{3 x}{7}) + 7 (- \frac{17}{7}) + 2}{3} + 5$

Rewrite the expression.

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 x + 7 (- \frac{17}{7}) + 2}{3} + 5$

Cancel the common factor of $7$ .

Move the leading negative in $- \frac{17}{7}$ into the numerator.

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 x + 7 (\frac{- 17}{7}) + 2}{3} + 5$

Cancel the common factor.

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 x + 7 (\frac{- 17}{7}) + 2}{3} + 5$

Rewrite the expression.

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 x - 17 + 2}{3} + 5$

Add $- 17$ and $2$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 x - 15}{3} + 5$

Factor $3$ out of $3 x - 15$ .

Factor $3$ out of $3 x$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 (x) - 15}{3} + 5$

Factor $3$ out of $- 15$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 x + 3 \cdot - 5}{3} + 5$

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

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 (x - 5)}{3} + 5$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{3 x}{7} - \frac{17}{7}) = \frac{3 (x - 5)}{3} + 5$

Divide $x - 5$ by $1$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = x - 5 + 5$

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

Add $- 5$ and $5$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = x + 0$

Add $x$ and $0$ .

$f (\frac{3 x}{7} - \frac{17}{7}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{3 x}{7} - \frac{17}{7}$ is the inverse of $f (x) = \frac{7 x + 2}{3} + 5$ .

$f^{- 1} (x) = \frac{3 x}{7} - \frac{17}{7}$

Find the Inverse f(x)=(7x+2)/3+5

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