Solve 5/xleq3-x/x

$$\frac{ 5 }{ x } \leq \frac{ 3-x }{ x }$$

Solve for x

$$x\in [-2,0)$$

Solution Steps

Subtract \frac{3-x}{x} from both sides.

$$\frac{5}{x}-\frac{3-x}{x}\leq 0$$

Since \frac{5}{x} and \frac{3-x}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{5-\left(3-x\right)}{x}\leq 0$$

Do the multiplications in 5-\left(3-x\right).

$$\frac{5-3+x}{x}\leq 0$$

Combine like terms in 5-3+x.

$$\frac{2+x}{x}\leq 0$$

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

$$x+2\geq 0 x<0$$

The solution satisfying both inequalities is x\in \left[-2,0\right).

$$x\in [-2,0)$$

Consider the case when x+2\leq 0 and x is positive.

$$x+2\leq 0 x>0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in [-2,0)$$