Solve 10/3x-1/6/frac{1{2}-1/3x}geq0

$$\frac { \frac { 10 } { 3 } x - \frac { 1 } { 6 } } { \frac { 1 } { 2 } - \frac { 1 } { 3 } x } \geq 0$$

Solve for x

$$x\in [\frac{1}{20},\frac{3}{2})$$

Solution Steps

For the quotient to be ≥0, \frac{10x}{3}-\frac{1}{6} and -\frac{x}{3}+\frac{1}{2} have to be both ≤0 or both ≥0, and -\frac{x}{3}+\frac{1}{2} cannot be zero. Consider the case when \frac{10x}{3}-\frac{1}{6}\leq 0 and -\frac{x}{3}+\frac{1}{2} is negative.

$$\frac{10x}{3}-\frac{1}{6}\leq 0 -\frac{x}{3}+\frac{1}{2}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when \frac{10x}{3}-\frac{1}{6}\geq 0 and -\frac{x}{3}+\frac{1}{2} is positive.

$$\frac{10x}{3}-\frac{1}{6}\geq 0 -\frac{x}{3}+\frac{1}{2}>0$$

The solution satisfying both inequalities is x\in \left[\frac{1}{20},\frac{3}{2}\right).

$$x\in [\frac{1}{20},\frac{3}{2})$$

The final solution is the union of the obtained solutions.

$$x\in [\frac{1}{20},\frac{3}{2})$$