Solve 1/3+1/2xleq5/6x-5/3

$$\frac { 1 } { 3 } + \frac { 1 } { 2 } x \leq \frac { 5 } { 6 } x - \frac { 5 } { 3 }$$

Solve for x

$$x\geq 6$$

Solution Steps

Subtract \frac{5}{6}x from both sides.

$$\frac{1}{3}+\frac{1}{2}x-\frac{5}{6}x\leq -\frac{5}{3}$$

Combine \frac{1}{2}x and -\frac{5}{6}x to get -\frac{1}{3}x.

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

Subtract \frac{1}{3} from both sides.

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

Since -\frac{5}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.

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

Subtract 1 from -5 to get -6.

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

Divide -6 by 3 to get -2.

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

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

$$x\geq -2\left(-3\right)$$

Multiply -2 and -3 to get 6.

$$x\geq 6$$