Solve for j 4(3j-2)<12j+16

$4 (3 j - 2) < 12 j + 16$

Simplify $4 (3 j - 2)$ .

Rewrite.

$0 + 0 + 4 (3 j - 2) < 12 j + 16$

Simplify by adding zeros.

$4 (3 j - 2) < 12 j + 16$

Apply the distributive property.

$4 (3 j) + 4 \cdot - 2 < 12 j + 16$

Multiply.

Multiply $3$ by $4$ .

$12 j + 4 \cdot - 2 < 12 j + 16$

Multiply $4$ by $- 2$ .

$12 j - 8 < 12 j + 16$

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

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

$12 j - 8 - 12 j < 16$

Combine the opposite terms in $12 j - 8 - 12 j$ .

Subtract $12 j$ from $12 j$ .

$0 - 8 < 16$

Subtract $8$ from $0$ .

$- 8 < 16$

Since $- 8 < 16$ , the inequality will always be true for any value of $j$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$