Algebra Examples

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\frac{2 - y}{3} = x$ .

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

Step 2.2

Multiply both sides by $3$ .

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

Step 2.3

Simplify.

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Step 2.3.1

Simplify the left side.

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Step 2.3.1.1

Simplify $\frac{2 - y}{3} \cdot 3$ .

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Step 2.3.1.1.1

Cancel the common factor of $3$ .

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Step 2.3.1.1.1.1

Cancel the common factor.

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

Step 2.3.1.1.1.2

Rewrite the expression.

$2 - y = x \cdot 3$

Step 2.3.1.1.2

Reorder $2$ and $- y$ .

$- y + 2 = x \cdot 3$

Step 2.3.2

Simplify the right side.

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Step 2.3.2.1

Move $3$ to the left of $x$ .

$- y + 2 = 3 x$

Step 2.4

Solve for $y$ .

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Step 2.4.1

Subtract $2$ from both sides of the equation.

$- y = 3 x - 2$

Step 2.4.2

Divide each term in $- y = 3 x - 2$ by $- 1$ and simplify.

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Step 2.4.2.1

Divide each term in $- y = 3 x - 2$ by $- 1$ .

$\frac{- y}{- 1} = \frac{3 x}{- 1} + \frac{- 2}{- 1}$

Step 2.4.2.2

Simplify the left side.

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Step 2.4.2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{3 x}{- 1} + \frac{- 2}{- 1}$

Step 2.4.2.2.2

Divide $y$ by $1$ .

$y = \frac{3 x}{- 1} + \frac{- 2}{- 1}$

Step 2.4.2.3

Simplify the right side.

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Step 2.4.2.3.1

Simplify each term.

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Step 2.4.2.3.1.1

Move the negative one from the denominator of $\frac{3 x}{- 1}$ .

$y = - 1 \cdot (3 x) + \frac{- 2}{- 1}$

Step 2.4.2.3.1.2

Rewrite $- 1 \cdot (3 x)$ as $- (3 x)$ .

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

Step 2.4.2.3.1.3

Multiply $3$ by $- 1$ .

$y = - 3 x + \frac{- 2}{- 1}$

Step 2.4.2.3.1.4

Divide $- 2$ by $- 1$ .

$y = - 3 x + 2$

Step 3

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

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

Step 4

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

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

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

Step 4.2.3

Simplify each term.

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Step 4.2.3.1

Cancel the common factor of $3$ .

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Step 4.2.3.1.1

Factor $3$ out of $- 3$ .

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

Step 4.2.3.1.2

Cancel the common factor.

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

Step 4.2.3.1.3

Rewrite the expression.

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

Step 4.2.3.2

Apply the distributive property.

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

Step 4.2.3.3

Multiply $- 1$ by $2$ .

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

Step 4.2.3.4

Multiply $- 1 (- x)$ .

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Step 4.2.3.4.1

Multiply $- 1$ by $- 1$ .

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

Step 4.2.3.4.2

Multiply $x$ by $1$ .

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

Step 4.2.4

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

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Step 4.2.4.1

Add $- 2$ and $2$ .

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

Step 4.2.4.2

Add $x$ and $0$ .

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

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

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

Step 4.3.3

Simplify the numerator.

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Step 4.3.3.1

Apply the distributive property.

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

Step 4.3.3.2

Multiply $- 3$ by $- 1$ .

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

Step 4.3.3.3

Multiply $- 1$ by $2$ .

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

Step 4.3.3.4

Subtract $2$ from $2$ .

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

Step 4.3.3.5

Add $3 x$ and $0$ .

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

Step 4.3.4

Cancel the common factor of $3$ .

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Step 4.3.4.1

Cancel the common factor.

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

Step 4.3.4.2

Divide $x$ by $1$ .

$f (- 3 x + 2) = x$

Step 4.4

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

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