Solve 2+2x-1/3leqx

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

Solve for x

$$x\geq 5$$

Solution Steps

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

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

Convert 2 to fraction \frac{6}{3}.

$$\frac{6}{3}+\frac{2}{3}x-\frac{1}{3}\leq x$$

Since \frac{6}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{6-1}{3}+\frac{2}{3}x\leq x$$

Subtract 1 from 6 to get 5.

$$\frac{5}{3}+\frac{2}{3}x\leq x$$

Subtract x from both sides.

$$\frac{5}{3}+\frac{2}{3}x-x\leq 0$$

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

$$\frac{5}{3}-\frac{1}{3}x\leq 0$$

Subtract \frac{5}{3} from both sides. Anything subtracted from zero gives its negation.

$$-\frac{1}{3}x\leq -\frac{5}{3}$$

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

$$x\geq -\frac{5}{3}\left(-3\right)$$

Express -\frac{5}{3}\left(-3\right) as a single fraction.

$$x\geq \frac{-5\left(-3\right)}{3}$$

Multiply -5 and -3 to get 15.

$$x\geq \frac{15}{3}$$

Divide 15 by 3 to get 5.

$$x\geq 5$$