Solve the Inequality for b -2(10-3b)>=6(b-3)-2

$- 2 (10 - 3 b) \geq 6 (b - 3) - 2$

Simplify $- 2 (10 - 3 b)$ .

Rewrite.

$0 + 0 - 2 (10 - 3 b) \geq 6 (b - 3) - 2$

Simplify by adding zeros.

$- 2 (10 - 3 b) \geq 6 (b - 3) - 2$

Apply the distributive property.

$- 2 \cdot 10 - 2 (- 3 b) \geq 6 (b - 3) - 2$

Multiply.

Multiply $- 2$ by $10$ .

$- 20 - 2 (- 3 b) \geq 6 (b - 3) - 2$

Multiply $- 3$ by $- 2$ .

$- 20 + 6 b \geq 6 (b - 3) - 2$

Simplify $6 (b - 3) - 2$ .

Simplify each term.

Apply the distributive property.

$- 20 + 6 b \geq 6 b + 6 \cdot - 3 - 2$

Multiply $6$ by $- 3$ .

$- 20 + 6 b \geq 6 b - 18 - 2$

Subtract $2$ from $- 18$ .

$- 20 + 6 b \geq 6 b - 20$

Move all terms containing $b$ to the left side of the inequality.

Subtract $6 b$ from both sides of the inequality.

$- 20 + 6 b - 6 b \geq - 20$

Combine the opposite terms in $- 20 + 6 b - 6 b$ .

Subtract $6 b$ from $6 b$ .

$- 20 + 0 \geq - 20$

Add $- 20$ and $0$ .

$- 20 \geq - 20$

Since $- 20 = - 20$ , the equation will always be true.

Always true

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$