Algebra Examples

$- 3 (4 - 2 v) = 6 v - 12$

Step 1

Simplify $- 3 (4 - 2 v)$ .

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Step 1.1

Rewrite.

$0 + 0 - 3 (4 - 2 v) = 6 v - 12$

Step 1.2

Simplify by adding zeros.

$- 3 (4 - 2 v) = 6 v - 12$

Step 1.3

Apply the distributive property.

$- 3 \cdot 4 - 3 (- 2 v) = 6 v - 12$

Step 1.4

Multiply.

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Step 1.4.1

Multiply $- 3$ by $4$ .

$- 12 - 3 (- 2 v) = 6 v - 12$

Step 1.4.2

Multiply $- 2$ by $- 3$ .

$- 12 + 6 v = 6 v - 12$

Step 2

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

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Step 2.1

Subtract $6 v$ from both sides of the equation.

$- 12 + 6 v - 6 v = - 12$

Step 2.2

Combine the opposite terms in $- 12 + 6 v - 6 v$ .

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Step 2.2.1

Subtract $6 v$ from $6 v$ .

$- 12 + 0 = - 12$

Step 2.2.2

Add $- 12$ and $0$ .

$- 12 = - 12$

Step 3

Since $- 12 = - 12$ , the equation will always be true for any value of $v$ .

All real numbers

Step 4

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$