Find the Inverse 2x-6y=1

$2 x - 6 y = 1$

Step 1

Subtract $2 x$ from both sides of the equation.

$- 6 y = 1 - 2 x$

Step 2

Divide each term in $- 6 y = 1 - 2 x$ by $- 6$ and simplify.

Divide each term in $- 6 y = 1 - 2 x$ by $- 6$ .

$\frac{- 6 y}{- 6} = \frac{1}{- 6} + \frac{- 2 x}{- 6}$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 y}{- 6} = \frac{1}{- 6} + \frac{- 2 x}{- 6}$

Divide $y$ by $1$ .

$y = \frac{1}{- 6} + \frac{- 2 x}{- 6}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y = - \frac{1}{6} + \frac{- 2 x}{- 6}$

Cancel the common factor of $- 2$ and $- 6$ .

Factor $- 2$ out of $- 2 x$ .

$y = - \frac{1}{6} + \frac{- 2 (x)}{- 6}$

Cancel the common factors.

Factor $- 2$ out of $- 6$ .

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

Cancel the common factor.

$y = - \frac{1}{6} + \frac{- 2 x}{- 2 \cdot 3}$

Rewrite the expression.

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

Step 3

Interchange the variables.

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

Step 4

Solve for $y$ .

Rewrite the equation as $- \frac{1}{6} + \frac{y}{3} = x$ .

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

Add $\frac{1}{6}$ to both sides of the equation.

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

Multiply both sides of the equation by $3$ .

$3 \frac{y}{3} = 3 (x + \frac{1}{6})$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$3 \frac{y}{3} = 3 (x + \frac{1}{6})$

Rewrite the expression.

$y = 3 (x + \frac{1}{6})$

Simplify the right side.

Simplify $3 (x + \frac{1}{6})$ .

Apply the distributive property.

$y = 3 x + 3 (\frac{1}{6})$

Cancel the common factor of $3$ .

Factor $3$ out of $6$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 5

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

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

Step 6

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{1}{6}$ into the numerator.

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

Factor $3$ out of $6$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Combine the opposite terms in $- \frac{1}{2} + x + \frac{1}{2}$ .

Add $- \frac{1}{2}$ and $\frac{1}{2}$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Simplify the numerator.

To write $3 x$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Combine $3 x$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

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

Multiply $2$ by $3$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $\frac{6 x + 1}{2} \cdot \frac{1}{3}$ .

Multiply $\frac{6 x + 1}{2}$ by $\frac{1}{3}$ .

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

Multiply $2$ by $3$ .

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

Simplify terms.

Combine the numerators over the common denominator.

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

Combine the opposite terms in $- 1 + 6 x + 1$ .

Add $- 1$ and $1$ .

$f (3 x + \frac{1}{2}) = \frac{6 x + 0}{6}$

Add $6 x$ and $0$ .

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

Cancel the common factor of $6$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse 2x-6y=1

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