Algebra Examples

y = 2 x - 1

$y = 2 x - 1$

Step 1

Interchange the variables.

x = 2 y - 1

$x = 2 y - 1$

Step 2

Solve for

y

$y$ .

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Step 2.1

Rewrite the equation as

2 y - 1 = x

$2 y - 1 = x$ .

2 y - 1 = x

$2 y - 1 = x$

Step 2.2

Add

1

$1$ to both sides of the equation.

2 y = x + 1

$2 y = x + 1$

Step 2.3

Divide each term in

2 y = x + 1

$2 y = x + 1$ by

2

$2$ and simplify.

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Step 2.3.1

Divide each term in

2 y = x + 1

$2 y = x + 1$ by

2

$2$ .

\frac{2 y}{2} = \frac{x}{2} + \frac{1}{2}

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

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of

2

$2$ .

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Step 2.3.2.1.1

Cancel the common factor.

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

Step 2.3.2.1.2

Divide

$y$ by

$1$ .

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

Step 3

Replace

$y$ with

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

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

Step 4

Verify if

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

$f (x) = 2 x - 1$ .

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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

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

$f$ into

$f^{- 1}$ .

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

Step 4.2.3

Combine the numerators over the common denominator.

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

Step 4.2.4

Combine the opposite terms in

$2 x - 1 + 1$ .

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Step 4.2.4.1

Add

$- 1$ and

$1$ .

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

Step 4.2.4.2

Add

$2 x$ and

$0$ .

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

Step 4.2.5

Cancel the common factor of

$2$ .

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Step 4.2.5.1

Cancel the common factor.

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

Step 4.2.5.2

Divide

$x$ by

$1$ .

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

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

$f (\frac{x}{2} + \frac{1}{2})$ by substituting in the value of

$f^{- 1}$ into

$f$ .

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

Step 4.3.3

Simplify each term.

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Step 4.3.3.1

Apply the distributive property.

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

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.

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

Step 4.3.3.2.2

Rewrite the expression.

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

Step 4.3.3.3

Cancel the common factor of

$2$ .

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Step 4.3.3.3.1

Cancel the common factor.

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

Step 4.3.3.3.2

Rewrite the expression.

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

Step 4.3.4

Combine the opposite terms in

$x + 1 - 1$ .

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Step 4.3.4.1

Subtract

$1$ from

$1$ .

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

Step 4.3.4.2

Add

$x$ and

$0$ .

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

Step 4.4

Since

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

$f (f^{- 1} (x)) = x$ , then

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

$f (x) = 2 x - 1$ .

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