Solve the Inequality for x 3(x+1)<=3x+11

$3 (x + 1) \leq 3 x + 11$

Simplify $3 (x + 1)$ .

Rewrite.

$0 + 0 + 3 (x + 1) \leq 3 x + 11$

Simplify by adding zeros.

$3 (x + 1) \leq 3 x + 11$

Apply the distributive property.

$3 x + 3 \cdot 1 \leq 3 x + 11$

Multiply $3$ by $1$ .

$3 x + 3 \leq 3 x + 11$

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

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

$3 x + 3 - 3 x \leq 11$

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

Subtract $3 x$ from $3 x$ .

$0 + 3 \leq 11$

Add $0$ and $3$ .

$3 \leq 11$

Since $3 \leq 11$ , the inequality will always be true.

Always true

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$