Solve for b 3(4b-5)<-28+12b

$3 (4 b - 5) < - 28 + 12 b$

Step 1

Simplify $3 (4 b - 5)$ .

Rewrite.

$0 + 0 + 3 (4 b - 5) < - 28 + 12 b$

Simplify by adding zeros.

$3 (4 b - 5) < - 28 + 12 b$

Apply the distributive property.

$3 (4 b) + 3 \cdot - 5 < - 28 + 12 b$

Multiply.

Multiply $4$ by $3$ .

$12 b + 3 \cdot - 5 < - 28 + 12 b$

Multiply $3$ by $- 5$ .

$12 b - 15 < - 28 + 12 b$

Step 2

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

Subtract $12 b$ from both sides of the inequality.

$12 b - 15 - 12 b < - 28$

Combine the opposite terms in $12 b - 15 - 12 b$ .

Subtract $12 b$ from $12 b$ .

$0 - 15 < - 28$

Subtract $15$ from $0$ .

$- 15 < - 28$

Step 3

Since $- 15 > - 28$ , there are no solutions.

No solution