Solve 1+2x/3leqx

$$\frac{ 1+2x }{ 3 } \leq x$$

Solve for x

$$x\geq 1$$

Solution Steps

Divide each term of 1+2x by 3 to get \frac{1}{3}+\frac{2}{3}x.

$$\frac{1}{3}+\frac{2}{3}x\leq x$$

Subtract x from both sides.

$$\frac{1}{3}+\frac{2}{3}x-x\leq 0$$

Combine \frac{2}{3}x and -x to get -\frac{1}{3}x.

$$\frac{1}{3}-\frac{1}{3}x\leq 0$$

Subtract \frac{1}{3} from both sides. Anything subtracted from zero gives its negation.

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

Multiply both sides by -3, the reciprocal of -\frac{1}{3}. Since -\frac{1}{3} is negative, the inequality direction is changed.

$$x\geq -\frac{1}{3}\left(-3\right)$$

Express -\frac{1}{3}\left(-3\right) as a single fraction.

$$x\geq \frac{-\left(-3\right)}{3}$$

Multiply -1 and -3 to get 3.

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

Divide 3 by 3 to get 1.

$$x\geq 1$$