Solve (-1/5)^-1*(x-2)/x+4leq0

$$\frac { ( - \frac { 1 } { 5 } ) ^ { - 1 } \cdot ( x - 2 ) } { x + 4 } \leq 0$$

Solve for x

$$x\in (-\infty,-4)\cup [2,\infty)$$

Solution Steps

Calculate -\frac{1}{5} to the power of -1 and get -5.

$$\frac{-5\left(x-2\right)}{x+4}\leq 0$$

Use the distributive property to multiply -5 by x-2.

$$\frac{-5x+10}{x+4}\leq 0$$

For the quotient to be ≤0, one of the values 10-5x and x+4 has to be ≥0, the other has to be ≤0, and x+4 cannot be zero. Consider the case when 10-5x\geq 0 and x+4 is negative.

$$10-5x\geq 0 x+4<0$$

The solution satisfying both inequalities is x<-4.

$$x<-4$$

Consider the case when 10-5x\leq 0 and x+4 is positive.

$$10-5x\leq 0 x+4>0$$

The solution satisfying both inequalities is x\geq 2.

$$x\geq 2$$

The final solution is the union of the obtained solutions.

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