Solve 4-5x/1+2xleq0

$$\frac { 4 - 5 x } { 1 + 2 x } \leq 0$$

Solve for x

$$x\in (-\infty,-\frac{1}{2})\cup [\frac{4}{5},\infty)$$

Solution Steps

For the quotient to be ≤0, one of the values 4-5x and 2x+1 has to be ≥0, the other has to be ≤0, and 2x+1 cannot be zero. Consider the case when 4-5x\geq 0 and 2x+1 is negative.

$$4-5x\geq 0 2x+1<0$$

The solution satisfying both inequalities is x<-\frac{1}{2}.

$$x<-\frac{1}{2}$$

Consider the case when 4-5x\leq 0 and 2x+1 is positive.

$$4-5x\leq 0 2x+1>0$$

The solution satisfying both inequalities is x\geq \frac{4}{5}.

$$x\geq \frac{4}{5}$$

The final solution is the union of the obtained solutions.

$$x<-\frac{1}{2}\text{; }x\geq \frac{4}{5}$$