Algebra Examples

$y = 10 x - 2$

Step 1

Interchange the variables.

$x = 10 y - 2$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $10 y - 2 = x$ .

$10 y - 2 = x$

Step 2.2

Add $2$ to both sides of the equation.

$10 y = x + 2$

Step 2.3

Divide each term in $10 y = x + 2$ by $10$ and simplify.

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Step 2.3.1

Divide each term in $10 y = x + 2$ by $10$ .

$\frac{10 y}{10} = \frac{x}{10} + \frac{2}{10}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $10$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{10 y}{10} = \frac{x}{10} + \frac{2}{10}$

Step 2.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{x}{10} + \frac{2}{10}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Cancel the common factor of $2$ and $10$ .

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Step 2.3.3.1.1

Factor $2$ out of $2$ .

$y = \frac{x}{10} + \frac{2 (1)}{10}$

Step 2.3.3.1.2

Cancel the common factors.

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Step 2.3.3.1.2.1

Factor $2$ out of $10$ .

$y = \frac{x}{10} + \frac{2 \cdot 1}{2 \cdot 5}$

Step 2.3.3.1.2.2

Cancel the common factor.

$y = \frac{x}{10} + \frac{2 \cdot 1}{2 \cdot 5}$

Step 2.3.3.1.2.3

Rewrite the expression.

$y = \frac{x}{10} + \frac{1}{5}$

Step 3

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

$f^{- 1} (x) = \frac{x}{10} + \frac{1}{5}$

Step 4

Verify if $f^{- 1} (x) = \frac{x}{10} + \frac{1}{5}$ is the inverse of $f (x) = 10 x - 2$ .

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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 $f^{- 1} (10 x - 2)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (10 x - 2) = \frac{10 x - 2}{10} + \frac{1}{5}$

Step 4.2.3

Cancel the common factor of $10 x - 2$ and $10$ .

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Step 4.2.3.1

Factor $2$ out of $10 x$ .

$f^{- 1} (10 x - 2) = \frac{2 (5 x) - 2}{10} + \frac{1}{5}$

Step 4.2.3.2

Factor $2$ out of $- 2$ .

$f^{- 1} (10 x - 2) = \frac{2 (5 x) + 2 (- 1)}{10} + \frac{1}{5}$

Step 4.2.3.3

Factor $2$ out of $2 (5 x) + 2 (- 1)$ .

$f^{- 1} (10 x - 2) = \frac{2 (5 x - 1)}{10} + \frac{1}{5}$

Step 4.2.3.4

Cancel the common factors.

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Step 4.2.3.4.1

Factor $2$ out of $10$ .

$f^{- 1} (10 x - 2) = \frac{2 (5 x - 1)}{2 (5)} + \frac{1}{5}$

Step 4.2.3.4.2

Cancel the common factor.

$f^{- 1} (10 x - 2) = \frac{2 (5 x - 1)}{2 \cdot 5} + \frac{1}{5}$

Step 4.2.3.4.3

Rewrite the expression.

$f^{- 1} (10 x - 2) = \frac{5 x - 1}{5} + \frac{1}{5}$

Step 4.2.4

Combine the numerators over the common denominator.

$f^{- 1} (10 x - 2) = \frac{5 x - 1 + 1}{5}$

Step 4.2.5

Combine the opposite terms in $5 x - 1 + 1$ .

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Step 4.2.5.1

Add $- 1$ and $1$ .

$f^{- 1} (10 x - 2) = \frac{5 x + 0}{5}$

Step 4.2.5.2

Add $5 x$ and $0$ .

$f^{- 1} (10 x - 2) = \frac{5 x}{5}$

Step 4.2.6

Cancel the common factor of $5$ .

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Step 4.2.6.1

Cancel the common factor.

$f^{- 1} (10 x - 2) = \frac{5 x}{5}$

Step 4.2.6.2

Divide $x$ by $1$ .

$f^{- 1} (10 x - 2) = x$

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 $f (\frac{x}{10} + \frac{1}{5})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x}{10} + \frac{1}{5}) = 10 (\frac{x}{10} + \frac{1}{5}) - 2$

Step 4.3.3

Simplify each term.

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Step 4.3.3.1

Apply the distributive property.

$f (\frac{x}{10} + \frac{1}{5}) = 10 (\frac{x}{10}) + 10 (\frac{1}{5}) - 2$

Step 4.3.3.2

Cancel the common factor of $10$ .

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Step 4.3.3.2.1

Cancel the common factor.

$f (\frac{x}{10} + \frac{1}{5}) = 10 (\frac{x}{10}) + 10 (\frac{1}{5}) - 2$

Step 4.3.3.2.2

Rewrite the expression.

$f (\frac{x}{10} + \frac{1}{5}) = x + 10 (\frac{1}{5}) - 2$

Step 4.3.3.3

Cancel the common factor of $5$ .

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Step 4.3.3.3.1

Factor $5$ out of $10$ .

$f (\frac{x}{10} + \frac{1}{5}) = x + 5 (2) (\frac{1}{5}) - 2$

Step 4.3.3.3.2

Cancel the common factor.

$f (\frac{x}{10} + \frac{1}{5}) = x + 5 \cdot (2 (\frac{1}{5})) - 2$

Step 4.3.3.3.3

Rewrite the expression.

$f (\frac{x}{10} + \frac{1}{5}) = x + 2 - 2$

Step 4.3.4

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

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Step 4.3.4.1

Subtract $2$ from $2$ .

$f (\frac{x}{10} + \frac{1}{5}) = x + 0$

Step 4.3.4.2

Add $x$ and $0$ .

$f (\frac{x}{10} + \frac{1}{5}) = x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x}{10} + \frac{1}{5}$ is the inverse of $f (x) = 10 x - 2$ .

$f^{- 1} (x) = \frac{x}{10} + \frac{1}{5}$