Solve the Inequality for n 3<=1/3+n

$3 \leq \frac{1}{3} + n$

Step 1

Rewrite so $n$ is on the left side of the inequality.

$\frac{1}{3} + n \geq 3$

Step 2

Move all terms not containing $n$ to the right side of the inequality.

Subtract $\frac{1}{3}$ from both sides of the inequality.

$n \geq 3 - \frac{1}{3}$

To write $3$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$n \geq 3 \cdot \frac{3}{3} - \frac{1}{3}$

Combine $3$ and $\frac{3}{3}$ .

$n \geq \frac{3 \cdot 3}{3} - \frac{1}{3}$

Combine the numerators over the common denominator.

$n \geq \frac{3 \cdot 3 - 1}{3}$

Simplify the numerator.

Multiply $3$ by $3$ .

$n \geq \frac{9 - 1}{3}$

Subtract $1$ from $9$ .

$n \geq \frac{8}{3}$

Step 3

The result can be shown in multiple forms.

Inequality Form:

$n \geq \frac{8}{3}$

Interval Notation:

$[\frac{8}{3}, \infty)$