Solve 2x-3/2-xgeq0

$$\frac { 2 x - 3 } { 2 - x } \geq 0$$

Solve for x

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

Solution Steps

For the quotient to be ≥0, 2x-3 and 2-x have to be both ≤0 or both ≥0, and 2-x cannot be zero. Consider the case when 2x-3\leq 0 and 2-x is negative.

$$2x-3\leq 0 2-x<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when 2x-3\geq 0 and 2-x is positive.

$$2x-3\geq 0 2-x>0$$

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

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

The final solution is the union of the obtained solutions.

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