Algebra Examples

$- \frac{5}{2} (4 - 6 k) = - \frac{1}{3} (12 k + 3 + 3 k)$

Step 1

Since $k$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$- \frac{1}{3} \cdot (12 k + 3 + 3 k) = - \frac{5}{2} \cdot (4 - 6 k)$

Step 2

Simplify $- \frac{1}{3} \cdot (12 k + 3 + 3 k)$ .

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Step 2.1

Rewrite.

$0 + 0 - \frac{1}{3} \cdot (12 k + 3 + 3 k) = - \frac{5}{2} \cdot (4 - 6 k)$

Step 2.2

Simplify by adding zeros.

$- \frac{1}{3} \cdot (12 k + 3 + 3 k) = - \frac{5}{2} \cdot (4 - 6 k)$

Step 2.3

Add $12 k$ and $3 k$ .

$- \frac{1}{3} \cdot (15 k + 3) = - \frac{5}{2} \cdot (4 - 6 k)$

Step 2.4

Apply the distributive property.

$- \frac{1}{3} (15 k) - \frac{1}{3} \cdot 3 = - \frac{5}{2} \cdot (4 - 6 k)$

Step 2.5

Cancel the common factor of $3$ .

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Step 2.5.1

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$\frac{- 1}{3} (15 k) - \frac{1}{3} \cdot 3 = - \frac{5}{2} \cdot (4 - 6 k)$

Step 2.5.2

Factor $3$ out of $15 k$ .

$\frac{- 1}{3} (3 (5 k)) - \frac{1}{3} \cdot 3 = - \frac{5}{2} \cdot (4 - 6 k)$

Step 2.5.3

Cancel the common factor.

$\frac{- 1}{3} (3 (5 k)) - \frac{1}{3} \cdot 3 = - \frac{5}{2} \cdot (4 - 6 k)$

Step 2.5.4

Rewrite the expression.

$- 1 (5 k) - \frac{1}{3} \cdot 3 = - \frac{5}{2} \cdot (4 - 6 k)$

Step 2.6

Multiply $5$ by $- 1$ .

$- 5 k - \frac{1}{3} \cdot 3 = - \frac{5}{2} \cdot (4 - 6 k)$

Step 2.7

Cancel the common factor of $3$ .

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Step 2.7.1

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$- 5 k + \frac{- 1}{3} \cdot 3 = - \frac{5}{2} \cdot (4 - 6 k)$

Step 2.7.2

Cancel the common factor.

$- 5 k + \frac{- 1}{3} \cdot 3 = - \frac{5}{2} \cdot (4 - 6 k)$

Step 2.7.3

Rewrite the expression.

$- 5 k - 1 = - \frac{5}{2} \cdot (4 - 6 k)$

Step 3

Simplify $- \frac{5}{2} \cdot (4 - 6 k)$ .

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Step 3.1

Apply the distributive property.

$- 5 k - 1 = - \frac{5}{2} \cdot 4 - \frac{5}{2} (- 6 k)$

Step 3.2

Cancel the common factor of $2$ .

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Step 3.2.1

Move the leading negative in $- \frac{5}{2}$ into the numerator.

$- 5 k - 1 = \frac{- 5}{2} \cdot 4 - \frac{5}{2} (- 6 k)$

Step 3.2.2

Factor $2$ out of $4$ .

$- 5 k - 1 = \frac{- 5}{2} \cdot (2 (2)) - \frac{5}{2} (- 6 k)$

Step 3.2.3

Cancel the common factor.

$- 5 k - 1 = \frac{- 5}{2} \cdot (2 \cdot 2) - \frac{5}{2} (- 6 k)$

Step 3.2.4

Rewrite the expression.

$- 5 k - 1 = - 5 \cdot 2 - \frac{5}{2} (- 6 k)$

Step 3.3

Multiply $- 5$ by $2$ .

$- 5 k - 1 = - 10 - \frac{5}{2} (- 6 k)$

Step 3.4

Cancel the common factor of $2$ .

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Step 3.4.1

Move the leading negative in $- \frac{5}{2}$ into the numerator.

$- 5 k - 1 = - 10 + \frac{- 5}{2} (- 6 k)$

Step 3.4.2

Factor $2$ out of $- 6 k$ .

$- 5 k - 1 = - 10 + \frac{- 5}{2} (2 (- 3 k))$

Step 3.4.3

Cancel the common factor.

$- 5 k - 1 = - 10 + \frac{- 5}{2} (2 (- 3 k))$

Step 3.4.4

Rewrite the expression.

$- 5 k - 1 = - 10 - 5 (- 3 k)$

Step 3.5

Multiply $- 3$ by $- 5$ .

$- 5 k - 1 = - 10 + 15 k$

Step 4

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

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Step 4.1

Subtract $15 k$ from both sides of the equation.

$- 5 k - 1 - 15 k = - 10$

Step 4.2

Subtract $15 k$ from $- 5 k$ .

$- 20 k - 1 = - 10$

Step 5

Move all terms not containing $k$ to the right side of the equation.

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Step 5.1

Add $1$ to both sides of the equation.

$- 20 k = - 10 + 1$

Step 5.2

Add $- 10$ and $1$ .

$- 20 k = - 9$

Step 6

Divide each term in $- 20 k = - 9$ by $- 20$ and simplify.

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Step 6.1

Divide each term in $- 20 k = - 9$ by $- 20$ .

$\frac{- 20 k}{- 20} = \frac{- 9}{- 20}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $- 20$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{- 20 k}{- 20} = \frac{- 9}{- 20}$

Step 6.2.1.2

Divide $k$ by $1$ .

$k = \frac{- 9}{- 20}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Dividing two negative values results in a positive value.

$k = \frac{9}{20}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$k = \frac{9}{20}$

Decimal Form:

$k = 0.45$