Solve the Inequality for g 8(3g-2)<=12(2g+1)

$8 (3 g - 2) \leq 12 (2 g + 1)$

Simplify $8 (3 g - 2)$ .

Start simplifying.

$0 + 0 + 8 (3 g - 2) \leq 12 (2 g + 1)$

Simplify by adding zeros.

$8 (3 g - 2) \leq 12 (2 g + 1)$

Apply the distributive property.

$8 (3 g) + 8 \cdot - 2 \leq 12 (2 g + 1)$

Multiply.

Multiply $3$ by $8$ .

$24 g + 8 \cdot - 2 \leq 12 (2 g + 1)$

Multiply $8$ by $- 2$ .

$24 g - 16 \leq 12 (2 g + 1)$

Simplify $12 (2 g + 1)$ .

Apply the distributive property.

$24 g - 16 \leq 12 (2 g) + 12 \cdot 1$

Multiply.

Multiply $2$ by $12$ .

$24 g - 16 \leq 24 g + 12 \cdot 1$

Multiply $12$ by $1$ .

$24 g - 16 \leq 24 g + 12$

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

Subtract $24 g$ from both sides of the inequality.

$24 g - 16 - 24 g \leq 12$

Combine the opposite terms in $24 g - 16 - 24 g$ .

Subtract $24 g$ from $24 g$ .

$0 - 16 \leq 12$

Subtract $16$ from $0$ .

$- 16 \leq 12$

Since $- 16 \leq 12$ , the inequality will always be true.

Always true

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$