Find the Inverse y=(3x+1)/2

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{3 y + 1}{2} = x$ .

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

Multiply both sides by $2$ .

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

Simplify.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$3 y + 1 = x \cdot 2$

Simplify the right side.

Move $2$ to the left of $x$ .

$3 y + 1 = 2 x$

Solve for $y$ .

Subtract $1$ from both sides of the equation.

$3 y = 2 x - 1$

Divide each term in $3 y = 2 x - 1$ by $3$ and simplify.

Divide each term in $3 y = 2 x - 1$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 3

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

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

Step 4

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

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

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

Combine the numerators over the common denominator.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Combine the opposite terms in $3 x + 1 - 1$ .

Subtract $1$ from $1$ .

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

Add $3 x$ and $0$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Apply the distributive property.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{1}{3}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Add $- 1$ and $1$ .

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

Add $2 x$ and $0$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse y=(3x+1)/2

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