Solve 2/3-1/2xgeq1/6+x

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

Solve for x

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

Solution Steps

Subtract x from both sides.

$$\frac{2}{3}-\frac{1}{2}x-x\geq \frac{1}{6}$$

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

$$\frac{2}{3}-\frac{3}{2}x\geq \frac{1}{6}$$

Subtract \frac{2}{3} from both sides.

$$-\frac{3}{2}x\geq \frac{1}{6}-\frac{2}{3}$$

Least common multiple of 6 and 3 is 6. Convert \frac{1}{6} and \frac{2}{3} to fractions with denominator 6.

$$-\frac{3}{2}x\geq \frac{1}{6}-\frac{4}{6}$$

Since \frac{1}{6} and \frac{4}{6} have the same denominator, subtract them by subtracting their numerators.

$$-\frac{3}{2}x\geq \frac{1-4}{6}$$

Subtract 4 from 1 to get -3.

$$-\frac{3}{2}x\geq \frac{-3}{6}$$

Reduce the fraction \frac{-3}{6} to lowest terms by extracting and canceling out 3.

$$-\frac{3}{2}x\geq -\frac{1}{2}$$

Multiply both sides by -\frac{2}{3}, the reciprocal of -\frac{3}{2}. Since -\frac{3}{2} is negative, the inequality direction is changed.

$$x\leq -\frac{1}{2}\left(-\frac{2}{3}\right)$$

Multiply -\frac{1}{2} times -\frac{2}{3} by multiplying numerator times numerator and denominator times denominator.

$$x\leq \frac{-\left(-2\right)}{2\times 3}$$

Do the multiplications in the fraction \frac{-\left(-2\right)}{2\times 3}.

$$x\leq \frac{2}{6}$$

Reduce the fraction \frac{2}{6} to lowest terms by extracting and canceling out 2.

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