Solve the Inequality for b 3(b+2)-7>=2b-5+b

$3 (b + 2) - 7 \geq 2 b - 5 + b$

Step 1

Rewrite so $b$ is on the left side of the inequality.

$2 b - 5 + b \leq 3 (b + 2) - 7$

Step 2

Add $2 b$ and $b$ .

$3 b - 5 \leq 3 (b + 2) - 7$

Step 3

Simplify $3 (b + 2) - 7$ .

Simplify each term.

Apply the distributive property.

$3 b - 5 \leq 3 b + 3 \cdot 2 - 7$

Multiply $3$ by $2$ .

$3 b - 5 \leq 3 b + 6 - 7$

Subtract $7$ from $6$ .

$3 b - 5 \leq 3 b - 1$

Step 4

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

Subtract $3 b$ from both sides of the inequality.

$3 b - 5 - 3 b \leq - 1$

Combine the opposite terms in $3 b - 5 - 3 b$ .

Subtract $3 b$ from $3 b$ .

$0 - 5 \leq - 1$

Subtract $5$ from $0$ .

$- 5 \leq - 1$

Step 5

Since $- 5 \leq - 1$ , the inequality will always be true.

Always true

Step 6

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$