Find the Inverse (2x-5)/3

$\frac{2 x - 5}{3}$

Step 1

Interchange the variables.

$x = \frac{2 y - 5}{3}$

Step 2

Solve for $y$ .

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

$\frac{2 y - 5}{3} = x$

Multiply both sides by $3$ .

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

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$2 y - 5 = x \cdot 3$

Simplify the right side.

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

$2 y - 5 = 3 x$

Solve for $y$ .

Add $5$ to both sides of the equation.

$2 y = 3 x + 5$

Divide each term in $2 y = 3 x + 5$ by $2$ and simplify.

Divide each term in $2 y = 3 x + 5$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Step 3

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

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

Step 4

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

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

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

Combine the numerators over the common denominator.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Combine the opposite terms in $2 x - 5 + 5$ .

Add $- 5$ and $5$ .

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

Add $2 x$ and $0$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Apply the distributive property.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Subtract $5$ from $5$ .

$f (\frac{3 x}{2} + \frac{5}{2}) = \frac{3 x + 0}{3}$

Add $3 x$ and $0$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse (2x-5)/3

Related Questions

Support LCMGCF.com by disabling your adblocker.