Algebra Examples

$\frac{2 x - 2}{4}$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\frac{2 y - 2}{4} = x$ .

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

Step 2.2

Cancel the common factor of $2 y - 2$ and $4$ .

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Step 2.2.1

Factor $2$ out of $2 y$ .

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

Step 2.2.2

Factor $2$ out of $- 2$ .

$\frac{2 (y) + 2 (- 1)}{4} = x$

Step 2.2.3

Factor $2$ out of $2 (y) + 2 (- 1)$ .

$\frac{2 (y - 1)}{4} = x$

Step 2.2.4

Cancel the common factors.

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Step 2.2.4.1

Factor $2$ out of $4$ .

$\frac{2 (y - 1)}{2 (2)} = x$

Step 2.2.4.2

Cancel the common factor.

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

Step 2.2.4.3

Rewrite the expression.

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

Step 2.3

Multiply both sides by $2$ .

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

Step 2.4

Simplify.

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Step 2.4.1

Simplify the left side.

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Step 2.4.1.1

Cancel the common factor of $2$ .

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Step 2.4.1.1.1

Cancel the common factor.

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

Step 2.4.1.1.2

Rewrite the expression.

$y - 1 = x \cdot 2$

Step 2.4.2

Simplify the right side.

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Step 2.4.2.1

Move $2$ to the left of $x$ .

$y - 1 = 2 x$

Step 2.5

Add $1$ to both sides of the equation.

$y = 2 x + 1$

Step 3

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

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

Step 4

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

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

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

Step 4.2.3

Simplify each term.

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Step 4.2.3.1

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

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Step 4.2.3.1.1

Factor $2$ out of $2 x$ .

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

Step 4.2.3.1.2

Factor $2$ out of $- 2$ .

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

Step 4.2.3.1.3

Factor $2$ out of $2 (x) + 2 (- 1)$ .

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

Step 4.2.3.2

Cancel the common factor of $2$ .

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Step 4.2.3.2.1

Factor $2$ out of $4$ .

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

Step 4.2.3.2.2

Cancel the common factor.

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

Step 4.2.3.2.3

Rewrite the expression.

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

Step 4.2.3.3

Cancel the common factor of $2$ .

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Step 4.2.3.3.1

Cancel the common factor.

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

Step 4.2.3.3.2

Divide $x - 1$ by $1$ .

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

Step 4.2.4

Combine the opposite terms in $x - 1 + 1$ .

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Step 4.2.4.1

Add $- 1$ and $1$ .

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

Step 4.2.4.2

Add $x$ and $0$ .

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

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

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

Step 4.3.3

Cancel the common factor of $2 (2 x + 1) - 2$ and $4$ .

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Step 4.3.3.1

Factor $2$ out of $- 2$ .

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

Step 4.3.3.2

Factor $2$ out of $2 (2 x + 1) + 2 (- 1)$ .

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

Step 4.3.3.3

Cancel the common factors.

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Step 4.3.3.3.1

Factor $2$ out of $4$ .

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

Step 4.3.3.3.2

Cancel the common factor.

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

Step 4.3.3.3.3

Rewrite the expression.

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

Step 4.3.4

Simplify the numerator.

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Step 4.3.4.1

Subtract $1$ from $1$ .

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

Step 4.3.4.2

Add $2 x$ and $0$ .

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

Step 4.3.5

Cancel the common factor of $2$ .

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Step 4.3.5.1

Cancel the common factor.

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

Step 4.3.5.2

Divide $x$ by $1$ .

$f (2 x + 1) = x$

Step 4.4

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

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