Algebra Examples

$2 (3 g + 2) = \frac{1}{2} \cdot (12 g + 8)$

Step 1

Subtract $\frac{1}{2} \cdot (12 g + 8)$ from both sides of the equation.

$2 (3 g + 2) - \frac{1}{2} \cdot (12 g + 8) = 0$

Step 2

Simplify $2 (3 g + 2) - \frac{1}{2} \cdot (12 g + 8)$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Apply the distributive property.

$2 (3 g) + 2 \cdot 2 - \frac{1}{2} \cdot (12 g + 8) = 0$

Step 2.1.2

Multiply $3$ by $2$ .

$6 g + 2 \cdot 2 - \frac{1}{2} \cdot (12 g + 8) = 0$

Step 2.1.3

Multiply $2$ by $2$ .

$6 g + 4 - \frac{1}{2} \cdot (12 g + 8) = 0$

Step 2.1.4

Apply the distributive property.

$6 g + 4 - \frac{1}{2} (12 g) - \frac{1}{2} \cdot 8 = 0$

Step 2.1.5

Cancel the common factor of $2$ .

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Step 2.1.5.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$6 g + 4 + \frac{- 1}{2} (12 g) - \frac{1}{2} \cdot 8 = 0$

Step 2.1.5.2

Factor $2$ out of $12 g$ .

$6 g + 4 + \frac{- 1}{2} (2 (6 g)) - \frac{1}{2} \cdot 8 = 0$

Step 2.1.5.3

Cancel the common factor.

$6 g + 4 + \frac{- 1}{2} (2 (6 g)) - \frac{1}{2} \cdot 8 = 0$

Step 2.1.5.4

Rewrite the expression.

$6 g + 4 - 1 (6 g) - \frac{1}{2} \cdot 8 = 0$

Step 2.1.6

Multiply $6$ by $- 1$ .

$6 g + 4 - 6 g - \frac{1}{2} \cdot 8 = 0$

Step 2.1.7

Cancel the common factor of $2$ .

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Step 2.1.7.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$6 g + 4 - 6 g + \frac{- 1}{2} \cdot 8 = 0$

Step 2.1.7.2

Factor $2$ out of $8$ .

$6 g + 4 - 6 g + \frac{- 1}{2} \cdot (2 (4)) = 0$

Step 2.1.7.3

Cancel the common factor.

$6 g + 4 - 6 g + \frac{- 1}{2} \cdot (2 \cdot 4) = 0$

Step 2.1.7.4

Rewrite the expression.

$6 g + 4 - 6 g - 1 \cdot 4 = 0$

Step 2.1.8

Multiply $- 1$ by $4$ .

$6 g + 4 - 6 g - 4 = 0$

Step 2.2

Combine the opposite terms in $6 g + 4 - 6 g - 4$ .

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Step 2.2.1

Subtract $6 g$ from $6 g$ .

$0 + 4 - 4 = 0$

Step 2.2.2

Add $0$ and $4$ .

$4 - 4 = 0$

Step 2.2.3

Subtract $4$ from $4$ .

$0 = 0$

Step 3

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

Always true