Find the Inverse y=-3x+6/7

$y = - 3 x + \frac{6}{7}$

Step 1

Interchange the variables.

$x = - 3 y + \frac{6}{7}$

Step 2

Solve for $y$ .

Rewrite the equation as $- 3 y + \frac{6}{7} = x$ .

$- 3 y + \frac{6}{7} = x$

Subtract $\frac{6}{7}$ from both sides of the equation.

$- 3 y = x - \frac{6}{7}$

Divide each term in $- 3 y = x - \frac{6}{7}$ by $- 3$ and simplify.

Divide each term in $- 3 y = x - \frac{6}{7}$ by $- 3$ .

$\frac{- 3 y}{- 3} = \frac{x}{- 3} + \frac{- \frac{6}{7}}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 y}{- 3} = \frac{x}{- 3} + \frac{- \frac{6}{7}}{- 3}$

Divide $y$ by $1$ .

$y = \frac{x}{- 3} + \frac{- \frac{6}{7}}{- 3}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y = - \frac{x}{3} + \frac{- \frac{6}{7}}{- 3}$

Multiply the numerator by the reciprocal of the denominator.

$y = - \frac{x}{3} - \frac{6}{7} \cdot \frac{1}{- 3}$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{6}{7}$ into the numerator.

$y = - \frac{x}{3} + \frac{- 6}{7} \cdot \frac{1}{- 3}$

Factor $3$ out of $- 6$ .

$y = - \frac{x}{3} + \frac{3 (- 2)}{7} \cdot \frac{1}{- 3}$

Factor $3$ out of $- 3$ .

$y = - \frac{x}{3} + \frac{3 \cdot - 2}{7} \cdot \frac{1}{3 \cdot - 1}$

Cancel the common factor.

$y = - \frac{x}{3} + \frac{3 \cdot - 2}{7} \cdot \frac{1}{3 \cdot - 1}$

Rewrite the expression.

$y = - \frac{x}{3} + \frac{- 2}{7} \cdot \frac{1}{- 1}$

Multiply $\frac{- 2}{7}$ by $\frac{1}{- 1}$ .

$y = - \frac{x}{3} + \frac{- 2}{7 \cdot - 1}$

Multiply $7$ by $- 1$ .

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

Dividing two negative values results in a positive value.

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

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = - \frac{x}{3} + \frac{2}{7}$ is the inverse of $f (x) = - 3 x + \frac{6}{7}$ .

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 $f^{- 1} (- 3 x + \frac{6}{7})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (- 3 x + \frac{6}{7}) = - \frac{- 3 x + \frac{6}{7}}{3} + \frac{2}{7}$

Simplify each term.

Cancel the common factor of $- 3 x + \frac{6}{7}$ and $3$ .

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

$f^{- 1} (- 3 x + \frac{6}{7}) = - \frac{3 (- x) + \frac{6}{7}}{3} + \frac{2}{7}$

Factor $3$ out of $\frac{6}{7}$ .

$f^{- 1} (- 3 x + \frac{6}{7}) = - \frac{3 (- x) + 3 (\frac{2}{7})}{3} + \frac{2}{7}$

Factor $3$ out of $3 (- x) + 3 (\frac{2}{7})$ .

$f^{- 1} (- 3 x + \frac{6}{7}) = - \frac{3 (- x + \frac{2}{7})}{3} + \frac{2}{7}$

Cancel the common factors.

Factor $3$ out of $3$ .

$f^{- 1} (- 3 x + \frac{6}{7}) = - \frac{3 (- x + \frac{2}{7})}{3 (1)} + \frac{2}{7}$

Cancel the common factor.

$f^{- 1} (- 3 x + \frac{6}{7}) = - \frac{3 (- x + \frac{2}{7})}{3 \cdot 1} + \frac{2}{7}$

Rewrite the expression.

$f^{- 1} (- 3 x + \frac{6}{7}) = - \frac{- x + \frac{2}{7}}{1} + \frac{2}{7}$

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

$f^{- 1} (- 3 x + \frac{6}{7}) = - (- x + \frac{2}{7}) + \frac{2}{7}$

Apply the distributive property.

$f^{- 1} (- 3 x + \frac{6}{7}) = x - \frac{2}{7} + \frac{2}{7}$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

$f^{- 1} (- 3 x + \frac{6}{7}) = 1 x - \frac{2}{7} + \frac{2}{7}$

Multiply $x$ by $1$ .

$f^{- 1} (- 3 x + \frac{6}{7}) = x - \frac{2}{7} + \frac{2}{7}$

Combine the opposite terms in $x - \frac{2}{7} + \frac{2}{7}$ .

Add $- \frac{2}{7}$ and $\frac{2}{7}$ .

$f^{- 1} (- 3 x + \frac{6}{7}) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (- 3 x + \frac{6}{7}) = x$

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

Set up the composite result function.

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

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

$f (- \frac{x}{3} + \frac{2}{7}) = - 3 (- \frac{x}{3} + \frac{2}{7}) + \frac{6}{7}$

Simplify each term.

Apply the distributive property.

$f (- \frac{x}{3} + \frac{2}{7}) = - 3 (- \frac{x}{3}) - 3 (\frac{2}{7}) + \frac{6}{7}$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{x}{3}$ into the numerator.

$f (- \frac{x}{3} + \frac{2}{7}) = - 3 \frac{- x}{3} - 3 (\frac{2}{7}) + \frac{6}{7}$

Factor $3$ out of $- 3$ .

$f (- \frac{x}{3} + \frac{2}{7}) = 3 (- 1) (\frac{- x}{3}) - 3 (\frac{2}{7}) + \frac{6}{7}$

Cancel the common factor.

$f (- \frac{x}{3} + \frac{2}{7}) = 3 \cdot (- 1 \frac{- x}{3}) - 3 (\frac{2}{7}) + \frac{6}{7}$

Rewrite the expression.

$f (- \frac{x}{3} + \frac{2}{7}) = - 1 (- x) - 3 (\frac{2}{7}) + \frac{6}{7}$

Multiply $- 1$ by $- 1$ .

$f (- \frac{x}{3} + \frac{2}{7}) = 1 x - 3 (\frac{2}{7}) + \frac{6}{7}$

Multiply $x$ by $1$ .

$f (- \frac{x}{3} + \frac{2}{7}) = x - 3 (\frac{2}{7}) + \frac{6}{7}$

Multiply $- 3 (\frac{2}{7})$ .

Combine $- 3$ and $\frac{2}{7}$ .

$f (- \frac{x}{3} + \frac{2}{7}) = x + \frac{- 3 \cdot 2}{7} + \frac{6}{7}$

Multiply $- 3$ by $2$ .

$f (- \frac{x}{3} + \frac{2}{7}) = x + \frac{- 6}{7} + \frac{6}{7}$

Move the negative in front of the fraction.

$f (- \frac{x}{3} + \frac{2}{7}) = x - \frac{6}{7} + \frac{6}{7}$

Combine the opposite terms in $x - \frac{6}{7} + \frac{6}{7}$ .

Add $- \frac{6}{7}$ and $\frac{6}{7}$ .

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

Add $x$ and $0$ .

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{x}{3} + \frac{2}{7}$ is the inverse of $f (x) = - 3 x + \frac{6}{7}$ .

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

Find the Inverse y=-3x+6/7

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