Algebra Examples

\frac{1}{4} k = 3 (- \frac{1}{4} k + 3)

$\frac{1}{4} k = 3 (- \frac{1}{4} k + 3)$

Step 1

Simplify

\frac{1}{4} k

$\frac{1}{4} k$ .

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Step 1.1

Rewrite.

0 + 0 + \frac{1}{4} k = 3 (- \frac{1}{4} k + 3)

$0 + 0 + \frac{1}{4} k = 3 (- \frac{1}{4} k + 3)$

Step 1.2

Simplify by adding zeros.

\frac{1}{4} k = 3 (- \frac{1}{4} k + 3)

$\frac{1}{4} k = 3 (- \frac{1}{4} k + 3)$

Step 1.3

Combine

\frac{1}{4}

$\frac{1}{4}$ and

k

$k$ .

\frac{k}{4} = 3 (- \frac{1}{4} k + 3)

$\frac{k}{4} = 3 (- \frac{1}{4} k + 3)$

\frac{k}{4} = 3 (- \frac{1}{4} k + 3)

$\frac{k}{4} = 3 (- \frac{1}{4} k + 3)$

Step 2

Simplify

3 (- \frac{1}{4} k + 3)

$3 (- \frac{1}{4} k + 3)$ .

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Step 2.1

Combine

k

$k$ and

\frac{1}{4}

$\frac{1}{4}$ .

\frac{k}{4} = 3 (- \frac{k}{4} + 3)

$\frac{k}{4} = 3 (- \frac{k}{4} + 3)$

Step 2.2

Apply the distributive property.

\frac{k}{4} = 3 (- \frac{k}{4}) + 3 \cdot 3

$\frac{k}{4} = 3 (- \frac{k}{4}) + 3 \cdot 3$

Step 2.3

Multiply

3 (- \frac{k}{4})

$3 (- \frac{k}{4})$ .

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Step 2.3.1

Multiply

- 1

$- 1$ by

3

$3$ .

\frac{k}{4} = - 3 \frac{k}{4} + 3 \cdot 3

$\frac{k}{4} = - 3 \frac{k}{4} + 3 \cdot 3$

Step 2.3.2

Combine

- 3

$- 3$ and

\frac{k}{4}

$\frac{k}{4}$ .

\frac{k}{4} = \frac{- 3 k}{4} + 3 \cdot 3

$\frac{k}{4} = \frac{- 3 k}{4} + 3 \cdot 3$

\frac{k}{4} = \frac{- 3 k}{4} + 3 \cdot 3

$\frac{k}{4} = \frac{- 3 k}{4} + 3 \cdot 3$

Step 2.4

Multiply

3

$3$ by

3

$3$ .

\frac{k}{4} = \frac{- 3 k}{4} + 9

$\frac{k}{4} = \frac{- 3 k}{4} + 9$

Step 2.5

Move the negative in front of the fraction.

\frac{k}{4} = - \frac{3 k}{4} + 9

$\frac{k}{4} = - \frac{3 k}{4} + 9$

\frac{k}{4} = - \frac{3 k}{4} + 9

$\frac{k}{4} = - \frac{3 k}{4} + 9$

Step 3

Move all terms containing

k

$k$ to the left side of the equation.

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Step 3.1

Add

\frac{3 k}{4}

$\frac{3 k}{4}$ to both sides of the equation.

\frac{k}{4} + \frac{3 k}{4} = 9

$\frac{k}{4} + \frac{3 k}{4} = 9$

Step 3.2

Combine the numerators over the common denominator.

\frac{k + 3 k}{4} = 9

$\frac{k + 3 k}{4} = 9$

Step 3.3

Add

k

$k$ and

3 k

$3 k$ .

\frac{4 k}{4} = 9

$\frac{4 k}{4} = 9$

Step 3.4

Cancel the common factor of

4

$4$ .

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Step 3.4.1

Cancel the common factor.

\frac{4 k}{4} = 9

$\frac{4 k}{4} = 9$

Step 3.4.2

Divide

k

$k$ by

1

$1$ .

k = 9

$k = 9$

k = 9

$k = 9$

k = 9

$k = 9$

\frac{1}{4} k = 3 (- \frac{1}{4} k + 3)

$\frac{1}{4} k = 3 (- \frac{1}{4} k + 3)$

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[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$