Solve for x 3<=1/3+x

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

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

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

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

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $3$ by $3$ .

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

Subtract $1$ from $9$ .

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

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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