Solve 4x-7/2xgeq0

$$\frac{ 4x-7 }{ 2x } \geq 0$$

Solve for x

$$x\in (-\infty,0)\cup [\frac{7}{4},\infty)$$

Solution Steps

For the quotient to be ≥0, 4x-7 and 2x have to be both ≤0 or both ≥0, and 2x cannot be zero. Consider the case when 4x-7\leq 0 and 2x is negative.

$$4x-7\leq 0 2x<0$$

The solution satisfying both inequalities is x<0.

$$x<0$$

Consider the case when 4x-7\geq 0 and 2x is positive.

$$4x-7\geq 0 2x>0$$

The solution satisfying both inequalities is x\geq \frac{7}{4}.

$$x\geq \frac{7}{4}$$

The final solution is the union of the obtained solutions.

$$x<0\text{; }x\geq \frac{7}{4}$$