Solve the Inequality for x 3(x-2)>=3x-10

$3 (x - 2) \geq 3 x - 10$

Step 1

Simplify $3 (x - 2)$ .

Rewrite.

$0 + 0 + 3 (x - 2) \geq 3 x - 10$

Simplify by adding zeros.

$3 (x - 2) \geq 3 x - 10$

Apply the distributive property.

$3 x + 3 \cdot - 2 \geq 3 x - 10$

Multiply $3$ by $- 2$ .

$3 x - 6 \geq 3 x - 10$

Step 2

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

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

$3 x - 6 - 3 x \geq - 10$

Combine the opposite terms in $3 x - 6 - 3 x$ .

Subtract $3 x$ from $3 x$ .

$0 - 6 \geq - 10$

Subtract $6$ from $0$ .

$- 6 \geq - 10$

Step 3

Since $- 6 \geq - 10$ , the inequality will always be true.

Always true

Step 4

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$