Solve x+7/x-3geq0

$$\frac { x + 7 } { x - 3 } \geq 0$$

Solve for x

$$x\in (-\infty,-7]\cup (3,\infty)$$

Solution Steps

For the quotient to be ≥0, x+7 and x-3 have to be both ≤0 or both ≥0, and x-3 cannot be zero. Consider the case when x+7\leq 0 and x-3 is negative.

$$x+7\leq 0 x-3<0$$

The solution satisfying both inequalities is x\leq -7.

$$x\leq -7$$

Consider the case when x+7\geq 0 and x-3 is positive.

$$x+7\geq 0 x-3>0$$

The solution satisfying both inequalities is x>3.

$$x>3$$

The final solution is the union of the obtained solutions.

$$x\leq -7\text{; }x>3$$