Solve the Inequality for x 3(x+2)>x

$3 (x + 2) > x$

Step 1

Simplify $3 (x + 2)$ .

Rewrite.

$0 + 0 + 3 (x + 2) > x$

Simplify by adding zeros.

$3 (x + 2) > x$

Apply the distributive property.

$3 x + 3 \cdot 2 > x$

Multiply $3$ by $2$ .

$3 x + 6 > x$

Step 2

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

Subtract $x$ from both sides of the inequality.

$3 x + 6 - x > 0$

Subtract $x$ from $3 x$ .

$2 x + 6 > 0$

Step 3

Subtract $6$ from both sides of the inequality.

$2 x > - 6$

Step 4

Divide each term in $2 x > - 6$ by $2$ and simplify.

Divide each term in $2 x > - 6$ by $2$ .

$\frac{2 x}{2} > \frac{- 6}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} > \frac{- 6}{2}$

Divide $x$ by $1$ .

$x > \frac{- 6}{2}$

Simplify the right side.

Divide $- 6$ by $2$ .

$x > - 3$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x > - 3$

Interval Notation:

$(- 3, \infty)$