Solve for b 26+6b>=2(3b+4)

$26 + 6 b \geq 2 (3 b + 4)$

Step 1

Simplify $2 (3 b + 4)$ .

Rewrite.

$26 + 6 b \geq 0 + 0 + 2 (3 b + 4)$

Simplify by adding zeros.

$26 + 6 b \geq 2 (3 b + 4)$

Apply the distributive property.

$26 + 6 b \geq 2 (3 b) + 2 \cdot 4$

Multiply.

Multiply $3$ by $2$ .

$26 + 6 b \geq 6 b + 2 \cdot 4$

Multiply $2$ by $4$ .

$26 + 6 b \geq 6 b + 8$

Step 2

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

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

$26 + 6 b - 6 b \geq 8$

Combine the opposite terms in $26 + 6 b - 6 b$ .

Subtract $6 b$ from $6 b$ .

$26 + 0 \geq 8$

Add $26$ and $0$ .

$26 \geq 8$

Step 3

Since $26 \geq 8$ , the inequality will always be true for any value of $b$ .

All real numbers

Step 4

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$