Pre-Algebra Examples

$6 r - 2 (2 - r) = 4 (2 r - 1)$

Step 1

Simplify $6 r - 2 (2 - r)$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Apply the distributive property.

$6 r - 2 \cdot 2 - 2 (- r) = 4 (2 r - 1)$

Step 1.1.2

Multiply $- 2$ by $2$ .

$6 r - 4 - 2 (- r) = 4 (2 r - 1)$

Step 1.1.3

Multiply $- 1$ by $- 2$ .

$6 r - 4 + 2 r = 4 (2 r - 1)$

Step 1.2

Add $6 r$ and $2 r$ .

$8 r - 4 = 4 (2 r - 1)$

Step 2

Simplify $4 (2 r - 1)$ .

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Step 2.1

Apply the distributive property.

$8 r - 4 = 4 (2 r) + 4 \cdot - 1$

Step 2.2

Multiply.

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Step 2.2.1

Multiply $2$ by $4$ .

$8 r - 4 = 8 r + 4 \cdot - 1$

Step 2.2.2

Multiply $4$ by $- 1$ .

$8 r - 4 = 8 r - 4$

Step 3

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

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Step 3.1

Subtract $8 r$ from both sides of the equation.

$8 r - 4 - 8 r = - 4$

Step 3.2

Combine the opposite terms in $8 r - 4 - 8 r$ .

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Step 3.2.1

Subtract $8 r$ from $8 r$ .

$0 - 4 = - 4$

Step 3.2.2

Subtract $4$ from $0$ .

$- 4 = - 4$

Step 4

Since $- 4 = - 4$ , the equation will always be true for any value of $r$ .

All real numbers

Step 5

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$