Algebra Examples

$a = \frac{1}{3} f (r + k)$

Step 1

Solve the equation for $a$ .

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Remove parentheses.

$a = \frac{1}{3} \cdot (f (r + k))$

Remove parentheses.

$a = \frac{1}{3} \cdot (f (r + k))$

Simplify $\frac{1}{3} \cdot (f (r + k))$ .

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Apply the distributive property.

$a = \frac{1}{3} \cdot (f r + f k)$

Apply the distributive property.

$a = \frac{1}{3} (f r) + \frac{1}{3} (f k)$

Multiply $\frac{1}{3} (f r)$ .

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Combine $f$ and $\frac{1}{3}$ .

$a = \frac{f}{3} r + \frac{1}{3} (f k)$

Combine $\frac{f}{3}$ and $r$ .

$a = \frac{f r}{3} + \frac{1}{3} (f k)$

Multiply $\frac{1}{3} (f k)$ .

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Combine $f$ and $\frac{1}{3}$ .

$a = \frac{f r}{3} + \frac{f}{3} k$

Combine $\frac{f}{3}$ and $k$ .

$a = \frac{f r}{3} + \frac{f k}{3}$

Reorder $\frac{f r}{3}$ and $\frac{f k}{3}$ .

$a = \frac{f k}{3} + \frac{f r}{3}$

Step 2

Solve the equation for $f$ .

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Simplify $\frac{1}{3} \cdot (f (r + k))$ .

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Apply the distributive property.

$\frac{f k}{3} + \frac{f r}{3} = \frac{1}{3} \cdot (f r + f k)$

Apply the distributive property.

$\frac{f k}{3} + \frac{f r}{3} = \frac{1}{3} (f r) + \frac{1}{3} (f k)$

Multiply $\frac{1}{3} (f r)$ .

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Combine $f$ and $\frac{1}{3}$ .

$\frac{f k}{3} + \frac{f r}{3} = \frac{f}{3} r + \frac{1}{3} (f k)$

Combine $\frac{f}{3}$ and $r$ .

$\frac{f k}{3} + \frac{f r}{3} = \frac{f r}{3} + \frac{1}{3} (f k)$

Multiply $\frac{1}{3} (f k)$ .

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Combine $f$ and $\frac{1}{3}$ .

$\frac{f k}{3} + \frac{f r}{3} = \frac{f r}{3} + \frac{f}{3} k$

Combine $\frac{f}{3}$ and $k$ .

$\frac{f k}{3} + \frac{f r}{3} = \frac{f r}{3} + \frac{f k}{3}$

Move all terms containing $f$ to the left side of the equation.

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Subtract $\frac{f r}{3}$ from both sides of the equation.

$\frac{f k}{3} + \frac{f r}{3} - \frac{f r}{3} = \frac{f k}{3}$

Subtract $\frac{f k}{3}$ from both sides of the equation.

$\frac{f k}{3} + \frac{f r}{3} - \frac{f r}{3} - \frac{f k}{3} = 0$

Combine the opposite terms in $\frac{f k}{3} + \frac{f r}{3} - \frac{f r}{3} - \frac{f k}{3}$ .

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Subtract $\frac{f r}{3}$ from $\frac{f r}{3}$ .

$\frac{f k}{3} + 0 - \frac{f k}{3} = 0$

Add $\frac{f k}{3}$ and $0$ .

$\frac{f k}{3} - \frac{f k}{3} = 0$

Subtract $\frac{f k}{3}$ from $\frac{f k}{3}$ .

$0 = 0$

Since $0 = 0$ , the equation will always be true.

Always true