Find the Inverse y=(2x-5)/(3x)

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$3 y, 1$

The LCM of one and any expression is the expression.

$3 y$

Multiply each term in $\frac{2 y - 5}{3 y} = x$ by $3 y$ to eliminate the fractions.

Multiply each term in $\frac{2 y - 5}{3 y} = x$ by $3 y$ .

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

Simplify the left side.

Rewrite using the commutative property of multiplication.

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

Cancel the common factor of $3$ .

Factor $3$ out of $3 y$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $y$ .

Cancel the common factor.

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

Rewrite the expression.

$2 y - 5 = x (3 y)$

Simplify the right side.

Rewrite using the commutative property of multiplication.

$2 y - 5 = 3 x y$

Solve the equation.

Subtract $3 x y$ from both sides of the equation.

$2 y - 5 - 3 x y = 0$

Add $5$ to both sides of the equation.

$2 y - 3 x y = 5$

Factor $y$ out of $2 y - 3 x y$ .

Factor $y$ out of $2 y$ .

$y \cdot 2 - 3 x y = 5$

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

$y \cdot 2 + y (- 3 x) = 5$

Factor $y$ out of $y \cdot 2 + y (- 3 x)$ .

$y (2 - 3 x) = 5$

Divide each term in $y (2 - 3 x) = 5$ by $2 - 3 x$ and simplify.

Divide each term in $y (2 - 3 x) = 5$ by $2 - 3 x$ .

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

Simplify the left side.

Cancel the common factor of $2 - 3 x$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Step 3

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

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

Step 4

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

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

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

Simplify the denominator.

Cancel the common factor of $3$ .

Factor $3$ out of $- 3$ .

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

Factor $3$ out of $3 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Rewrite $- 1 \frac{2 x - 5}{x}$ as $- \frac{2 x - 5}{x}$ .

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

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

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

Combine the numerators over the common denominator.

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

Rewrite $\frac{2 x - (2 x - 5)}{x}$ in a factored form.

Apply the distributive property.

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

Multiply $2$ by $- 1$ .

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

Multiply $- 1$ by $- 5$ .

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

Subtract $2 x$ from $2 x$ .

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

Add $0$ and $5$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

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

Combine $2$ and $\frac{5}{2 - 3 x}$ .

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

Multiply $2$ by $5$ .

$f (\frac{5}{2 - 3 x}) = \frac{\frac{10}{2 - 3 x} - 5}{3 (\frac{5}{2 - 3 x})}$

To write $- 5$ as a fraction with a common denominator, multiply by $\frac{2 - 3 x}{2 - 3 x}$ .

$f (\frac{5}{2 - 3 x}) = \frac{\frac{10}{2 - 3 x} - 5 \cdot \frac{2 - 3 x}{2 - 3 x}}{3 (\frac{5}{2 - 3 x})}$

Combine $- 5$ and $\frac{2 - 3 x}{2 - 3 x}$ .

$f (\frac{5}{2 - 3 x}) = \frac{\frac{10}{2 - 3 x} + \frac{- 5 (2 - 3 x)}{2 - 3 x}}{3 (\frac{5}{2 - 3 x})}$

Combine the numerators over the common denominator.

$f (\frac{5}{2 - 3 x}) = \frac{\frac{10 - 5 (2 - 3 x)}{2 - 3 x}}{3 (\frac{5}{2 - 3 x})}$

Reorder terms.

$f (\frac{5}{2 - 3 x}) = \frac{\frac{10 - 5 (2 - 3 x)}{- 3 x + 2}}{3 (\frac{5}{2 - 3 x})}$

Rewrite $\frac{10 - 5 (2 - 3 x)}{- 3 x + 2}$ in a factored form.

Factor $5$ out of $10 - 5 (2 - 3 x)$ .

Factor $5$ out of $10$ .

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

Factor $5$ out of $- 5 (2 - 3 x)$ .

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

Factor $5$ out of $5 (2) + 5 (- (2 - 3 x))$ .

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

Apply the distributive property.

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

Multiply $- 1$ by $2$ .

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

Multiply $- 3$ by $- 1$ .

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

Subtract $2$ from $2$ .

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

Add $0$ and $3 x$ .

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

Multiply $5$ by $3$ .

$f (\frac{5}{2 - 3 x}) = \frac{\frac{15 x}{- 3 x + 2}}{3 (\frac{5}{2 - 3 x})}$

Combine fractions.

Combine $3$ and $\frac{5}{2 - 3 x}$ .

$f (\frac{5}{2 - 3 x}) = \frac{\frac{15 x}{- 3 x + 2}}{\frac{3 \cdot 5}{2 - 3 x}}$

Multiply $3$ by $5$ .

$f (\frac{5}{2 - 3 x}) = \frac{\frac{15 x}{- 3 x + 2}}{\frac{15}{2 - 3 x}}$

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{5}{2 - 3 x}) = \frac{15 x}{- 3 x + 2} \cdot \frac{2 - 3 x}{15}$

Cancel the common factor of $15$ .

Factor $15$ out of $15 x$ .

$f (\frac{5}{2 - 3 x}) = \frac{15 (x)}{- 3 x + 2} \cdot \frac{2 - 3 x}{15}$

Cancel the common factor.

$f (\frac{5}{2 - 3 x}) = \frac{15 x}{- 3 x + 2} \cdot \frac{2 - 3 x}{15}$

Rewrite the expression.

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

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

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

Cancel the common factor of $2 - 3 x$ and $- 3 x + 2$ .

Reorder terms.

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

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse y=(2x-5)/(3x)

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