Algebra Examples

$y = 2 x + 3$

Step 1

Interchange the variables.

$x = 2 y + 3$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $2 y + 3 = x$ .

$2 y + 3 = x$

Step 2.2

Subtract $3$ from both sides of the equation.

$2 y = x - 3$

Step 2.3

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

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Step 2.3.1

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

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

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.

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

Step 2.3.2.1.2

Divide $y$ by $1$ .

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

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Move the negative in front of the fraction.

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

Step 3

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

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

Step 4

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

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

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

Step 4.2.3

Combine the numerators over the common denominator.

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

Step 4.2.4

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

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Step 4.2.4.1

Subtract $3$ from $3$ .

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

Step 4.2.4.2

Add $2 x$ and $0$ .

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

Step 4.2.5

Cancel the common factor of $2$ .

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Step 4.2.5.1

Cancel the common factor.

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

Step 4.2.5.2

Divide $x$ by $1$ .

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

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

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

Step 4.3.3

Simplify each term.

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Step 4.3.3.1

Apply the distributive property.

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

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.

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

Step 4.3.3.2.2

Rewrite the expression.

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

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 $- \frac{3}{2}$ into the numerator.

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

Step 4.3.3.3.2

Cancel the common factor.

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

Step 4.3.3.3.3

Rewrite the expression.

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

Step 4.3.4

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

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Step 4.3.4.1

Add $- 3$ and $3$ .

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

Step 4.3.4.2

Add $x$ and $0$ .

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

Step 4.4

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

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