Find the Inverse f(x)=(6x-7)/5

$f (x) = \frac{6 x - 7}{5}$

Write $f (x) = \frac{6 x - 7}{5}$ as an equation.

$y = \frac{6 x - 7}{5}$

Interchange the variables.

$x = \frac{6 y - 7}{5}$

Solve for $y$ .

Rewrite the equation as $\frac{6 y - 7}{5} = x$ .

$\frac{6 y - 7}{5} = x$

Multiply both sides by $5$ .

$\frac{6 y - 7}{5} \cdot 5 = x \cdot 5$

Simplify.

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{6 y - 7}{5} \cdot 5 = x \cdot 5$

Rewrite the expression.

$6 y - 7 = x \cdot 5$

Simplify the right side.

Move $5$ to the left of $x$ .

$6 y - 7 = 5 x$

Solve for $y$ .

Add $7$ to both sides of the equation.

$6 y = 5 x + 7$

Divide each term in $6 y = 5 x + 7$ by $6$ and simplify.

Divide each term in $6 y = 5 x + 7$ by $6$ .

$\frac{6 y}{6} = \frac{5 x}{6} + \frac{7}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 y}{6} = \frac{5 x}{6} + \frac{7}{6}$

Divide $y$ by $1$ .

$y = \frac{5 x}{6} + \frac{7}{6}$

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

$f^{- 1} (x) = \frac{5 x}{6} + \frac{7}{6}$

Verify if $f^{- 1} (x) = \frac{5 x}{6} + \frac{7}{6}$ is the inverse of $f (x) = \frac{6 x - 7}{5}$ .

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 $f^{- 1} (\frac{6 x - 7}{5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{6 x - 7}{5}) = \frac{5 (\frac{6 x - 7}{5})}{6} + \frac{7}{6}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{6 x - 7}{5}) = \frac{5 (\frac{6 x - 7}{5}) + 7}{6}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f^{- 1} (\frac{6 x - 7}{5}) = \frac{5 (\frac{6 x - 7}{5}) + 7}{6}$

Rewrite the expression.

$f^{- 1} (\frac{6 x - 7}{5}) = \frac{6 x - 7 + 7}{6}$

Combine the opposite terms in $6 x - 7 + 7$ .

Add $- 7$ and $7$ .

$f^{- 1} (\frac{6 x - 7}{5}) = \frac{6 x + 0}{6}$

Add $6 x$ and $0$ .

$f^{- 1} (\frac{6 x - 7}{5}) = \frac{6 x}{6}$

Cancel the common factor of $6$ .

Cancel the common factor.

$f^{- 1} (\frac{6 x - 7}{5}) = \frac{6 x}{6}$

Divide $x$ by $1$ .

$f^{- 1} (\frac{6 x - 7}{5}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{5 x}{6} + \frac{7}{6})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{5 x}{6} + \frac{7}{6}) = \frac{6 (\frac{5 x}{6} + \frac{7}{6}) - 7}{5}$

Simplify the numerator.

Apply the distributive property.

$f (\frac{5 x}{6} + \frac{7}{6}) = \frac{6 (\frac{5 x}{6}) + 6 (\frac{7}{6}) - 7}{5}$

Cancel the common factor of $6$ .

Cancel the common factor.

$f (\frac{5 x}{6} + \frac{7}{6}) = \frac{6 (\frac{5 x}{6}) + 6 (\frac{7}{6}) - 7}{5}$

Rewrite the expression.

$f (\frac{5 x}{6} + \frac{7}{6}) = \frac{5 x + 6 (\frac{7}{6}) - 7}{5}$

Cancel the common factor of $6$ .

Cancel the common factor.

$f (\frac{5 x}{6} + \frac{7}{6}) = \frac{5 x + 6 (\frac{7}{6}) - 7}{5}$

Rewrite the expression.

$f (\frac{5 x}{6} + \frac{7}{6}) = \frac{5 x + 7 - 7}{5}$

Subtract $7$ from $7$ .

$f (\frac{5 x}{6} + \frac{7}{6}) = \frac{5 x + 0}{5}$

Add $5 x$ and $0$ .

$f (\frac{5 x}{6} + \frac{7}{6}) = \frac{5 x}{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (\frac{5 x}{6} + \frac{7}{6}) = \frac{5 x}{5}$

Divide $x$ by $1$ .

$f (\frac{5 x}{6} + \frac{7}{6}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{5 x}{6} + \frac{7}{6}$ is the inverse of $f (x) = \frac{6 x - 7}{5}$ .

$f^{- 1} (x) = \frac{5 x}{6} + \frac{7}{6}$

Find the Inverse f(x)=(6x-7)/5

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