Algebra Examples

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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Step 2.1

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

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

Step 2.2

Subtract $\frac{5}{2}$ from both sides of the equation.

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

Step 2.3

Divide each term in $2 y = x - \frac{5}{2}$ by $2$ and simplify.

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Step 2.3.1

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

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

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $2$ .

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Step 2.3.2.1.1

Cancel the common factor.

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

Step 2.3.2.1.2

Divide $y$ by $1$ .

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

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Simplify each term.

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Step 2.3.3.1.1

Multiply the numerator by the reciprocal of the denominator.

$y = \frac{x}{2} - \frac{5}{2} \cdot \frac{1}{2}$

Step 2.3.3.1.2

Multiply $- \frac{5}{2} \cdot \frac{1}{2}$ .

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Step 2.3.3.1.2.1

Multiply $\frac{1}{2}$ by $\frac{5}{2}$ .

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

Step 2.3.3.1.2.2

Multiply $2$ by $2$ .

$y = \frac{x}{2} - \frac{5}{4}$

Step 3

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

$f^{- 1} (x) = \frac{x}{2} - \frac{5}{4}$

Step 4

Verify if $f^{- 1} (x) = \frac{x}{2} - \frac{5}{4}$ is the inverse of $f (x) = 2 x + \frac{5}{2}$ .

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Step 4.1

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Step 4.2

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

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

$f^{- 1} (2 x + \frac{5}{2}) = \frac{2 x + \frac{5}{2}}{2} - \frac{5}{4}$

Step 4.2.3

Simplify each term.

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Step 4.2.3.1

Simplify the numerator.

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Step 4.2.3.1.1

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

$f^{- 1} (2 x + \frac{5}{2}) = \frac{2 x \cdot \frac{2}{2} + \frac{5}{2}}{2} - \frac{5}{4}$

Step 4.2.3.1.2

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

$f^{- 1} (2 x + \frac{5}{2}) = \frac{\frac{2 x \cdot 2}{2} + \frac{5}{2}}{2} - \frac{5}{4}$

Step 4.2.3.1.3

Combine the numerators over the common denominator.

$f^{- 1} (2 x + \frac{5}{2}) = \frac{\frac{2 x \cdot 2 + 5}{2}}{2} - \frac{5}{4}$

Step 4.2.3.1.4

Multiply $2$ by $2$ .

$f^{- 1} (2 x + \frac{5}{2}) = \frac{\frac{4 x + 5}{2}}{2} - \frac{5}{4}$

Step 4.2.3.2

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (2 x + \frac{5}{2}) = \frac{4 x + 5}{2} \cdot \frac{1}{2} - \frac{5}{4}$

Step 4.2.3.3

Multiply $\frac{4 x + 5}{2} \cdot \frac{1}{2}$ .

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Step 4.2.3.3.1

Multiply $\frac{4 x + 5}{2}$ by $\frac{1}{2}$ .

$f^{- 1} (2 x + \frac{5}{2}) = \frac{4 x + 5}{2 \cdot 2} - \frac{5}{4}$

Step 4.2.3.3.2

Multiply $2$ by $2$ .

$f^{- 1} (2 x + \frac{5}{2}) = \frac{4 x + 5}{4} - \frac{5}{4}$

Step 4.2.4

Simplify terms.

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Step 4.2.4.1

Combine the numerators over the common denominator.

$f^{- 1} (2 x + \frac{5}{2}) = \frac{4 x + 5 - 5}{4}$

Step 4.2.4.2

Combine the opposite terms in $4 x + 5 - 5$ .

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Step 4.2.4.2.1

Subtract $5$ from $5$ .

$f^{- 1} (2 x + \frac{5}{2}) = \frac{4 x + 0}{4}$

Step 4.2.4.2.2

Add $4 x$ and $0$ .

$f^{- 1} (2 x + \frac{5}{2}) = \frac{4 x}{4}$

Step 4.2.4.3

Cancel the common factor of $4$ .

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Step 4.2.4.3.1

Cancel the common factor.

$f^{- 1} (2 x + \frac{5}{2}) = \frac{4 x}{4}$

Step 4.2.4.3.2

Divide $x$ by $1$ .

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

Step 4.3

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

Evaluate $f (\frac{x}{2} - \frac{5}{4})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x}{2} - \frac{5}{4}) = 2 (\frac{x}{2} - \frac{5}{4}) + \frac{5}{2}$

Step 4.3.3

Simplify each term.

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Step 4.3.3.1

Apply the distributive property.

$f (\frac{x}{2} - \frac{5}{4}) = 2 (\frac{x}{2}) + 2 (- \frac{5}{4}) + \frac{5}{2}$

Step 4.3.3.2

Cancel the common factor of $2$ .

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Step 4.3.3.2.1

Cancel the common factor.

$f (\frac{x}{2} - \frac{5}{4}) = 2 (\frac{x}{2}) + 2 (- \frac{5}{4}) + \frac{5}{2}$

Step 4.3.3.2.2

Rewrite the expression.

$f (\frac{x}{2} - \frac{5}{4}) = x + 2 (- \frac{5}{4}) + \frac{5}{2}$

Step 4.3.3.3

Cancel the common factor of $2$ .

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Step 4.3.3.3.1

Move the leading negative in $- \frac{5}{4}$ into the numerator.

$f (\frac{x}{2} - \frac{5}{4}) = x + 2 (\frac{- 5}{4}) + \frac{5}{2}$

Step 4.3.3.3.2

Factor $2$ out of $4$ .

$f (\frac{x}{2} - \frac{5}{4}) = x + 2 (\frac{- 5}{2 (2)}) + \frac{5}{2}$

Step 4.3.3.3.3

Cancel the common factor.

$f (\frac{x}{2} - \frac{5}{4}) = x + 2 (\frac{- 5}{2 \cdot 2}) + \frac{5}{2}$

Step 4.3.3.3.4

Rewrite the expression.

$f (\frac{x}{2} - \frac{5}{4}) = x + \frac{- 5}{2} + \frac{5}{2}$

Step 4.3.3.4

Move the negative in front of the fraction.

$f (\frac{x}{2} - \frac{5}{4}) = x - \frac{5}{2} + \frac{5}{2}$

Step 4.3.4

Combine the opposite terms in $x - \frac{5}{2} + \frac{5}{2}$ .

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Step 4.3.4.1

Add $- \frac{5}{2}$ and $\frac{5}{2}$ .

$f (\frac{x}{2} - \frac{5}{4}) = x + 0$

Step 4.3.4.2

Add $x$ and $0$ .

$f (\frac{x}{2} - \frac{5}{4}) = x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x}{2} - \frac{5}{4}$ is the inverse of $f (x) = 2 x + \frac{5}{2}$ .

$f^{- 1} (x) = \frac{x}{2} - \frac{5}{4}$