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

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

Step 1

Write $F (x) = \frac{2 x - 3}{5}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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

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

Multiply both sides by $5$ .

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

Simplify.

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

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

Rewrite the expression.

$2 y - 3 = x \cdot 5$

Simplify the right side.

Move $5$ to the left of $x$ .

$2 y - 3 = 5 x$

Solve for $y$ .

Add $3$ to both sides of the equation.

$2 y = 5 x + 3$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Step 4

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

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

Step 5

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

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

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

Combine the numerators over the common denominator.

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Rewrite the expression.

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

Combine the opposite terms in $2 x - 3 + 3$ .

Add $- 3$ and $3$ .

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

Add $2 x$ and $0$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Apply the distributive property.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Subtract $3$ from $3$ .

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

Add $5 x$ and $0$ .

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

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

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