Solve for b 3(2-b)<10-3(b-6)

$3 (2 - b) < 10 - 3 (b - 6)$

Step 1

Simplify $3 (2 - b)$ .

Rewrite.

$0 + 0 + 3 (2 - b) < 10 - 3 (b - 6)$

Simplify by adding zeros.

$3 (2 - b) < 10 - 3 (b - 6)$

Apply the distributive property.

$3 \cdot 2 + 3 (- b) < 10 - 3 (b - 6)$

Multiply.

Multiply $3$ by $2$ .

$6 + 3 (- b) < 10 - 3 (b - 6)$

Multiply $- 1$ by $3$ .

$6 - 3 b < 10 - 3 (b - 6)$

Step 2

Simplify $10 - 3 (b - 6)$ .

Simplify each term.

Apply the distributive property.

$6 - 3 b < 10 - 3 b - 3 \cdot - 6$

Multiply $- 3$ by $- 6$ .

$6 - 3 b < 10 - 3 b + 18$

Add $10$ and $18$ .

$6 - 3 b < - 3 b + 28$

Step 3

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

Add $3 b$ to both sides of the inequality.

$6 - 3 b + 3 b < 28$

Combine the opposite terms in $6 - 3 b + 3 b$ .

Add $- 3 b$ and $3 b$ .

$6 + 0 < 28$

Add $6$ and $0$ .

$6 < 28$

Step 4

Since $6 < 28$ , the inequality will always be true for any value of $b$ .

All real numbers

Step 5

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$