To solve the inequality:

$\left| \frac{x - 3}{x + 6} \right| \geq 2$

We need to split this into two cases based on the definition of absolute value:

$\frac{x - 3}{x + 6} \geq 2 \quad \text{or} \quad \frac{x - 3}{x + 6} \leq -2$

Case 1: $\frac{x - 3}{x + 6} \geq 2$

1. Multiply both sides by $x + 6$, but note that $x + 6$ could be positive or negative. Therefore, we must consider two subcases:
   • If $x + 6 > 0$ (i.e., $x > -6$), multiply directly.
   • If $x + 6 < 0$ (i.e., $x < -6$), reverse the inequality.

Subcase 1.1: $x > -6$

$\frac{x - 3}{x + 6} \geq 2 \quad \Rightarrow \quad x - 3 \geq 2(x + 6)$ Expanding: $x - 3 \geq 2x + 12$ Simplifying: $x - 2x \geq 12 + 3 \quad \Rightarrow \quad -x \geq 15 \quad \Rightarrow \quad x \leq -15$

But since this must hold under the condition $x > -6$, there is no solution in this case, because $x \leq -15$ contradicts $x > -6$.

Subcase 1.2: $x < -6$

Now reverse the inequality: $x - 3 \leq 2(x + 6)$ Expanding: $x - 3 \leq 2x + 12$ Simplifying: $x - 2x \leq 12 + 3 \quad \Rightarrow \quad -x \leq 15 \quad \Rightarrow \quad x \geq -15$ This gives the range $-15 \leq x < -6$.

Case 2: $\frac{x - 3}{x + 6} \leq -2$

Again, multiply both sides by $x + 6$ and consider two subcases.

Subcase 2.1: $x > -6$

$\frac{x - 3}{x + 6} \leq -2 \quad \Rightarrow \quad x - 3 \leq -2(x + 6)$ Expanding: $x - 3 \leq -2x - 12$ Simplifying: $x + 2x \leq -12 + 3 \quad \Rightarrow \quad 3x \leq -9 \quad \Rightarrow \quad x \leq -3$ Thus, the solution for this subcase is $x \leq -3$, but under the condition $x > -6$, so the solution is $-6 < x \leq -3$.

Subcase 2.2: $x < -6$

Reverse the inequality: $x - 3 \geq -2(x + 6)$ Expanding: $x - 3 \geq -2x - 12$ Simplifying: $x + 2x \geq -12 + 3 \quad \Rightarrow \quad 3x \geq -9 \quad \Rightarrow \quad x \geq -3$ But since $x < -6$, this is a contradiction, so no solution exists in this subcase.

Final Solution:

Combining both cases, the solution is:

$-15 \leq x < -6 \quad \text{or} \quad -6 < x \leq -3$

Note that $x = -6$ is excluded because it would make the denominator zero.

Would you like further clarification or more details?

---

Here are 5 related questions for further practice:

1. Solve $\left| \frac{x + 2}{x - 4} \right| \geq 3$.
2. Solve $\left| \frac{2x - 5}{x + 1} \right| \leq 4$.
3. Solve the inequality $\frac{x - 1}{x + 3} > 1$.
4. Solve $\left| \frac{x + 3}{x - 7} \right| = 2$.
5. Solve the quadratic inequality $\left| x^2 - 3x + 2 \right| \leq 5$.

Tip: Always remember to consider the sign of the denominator when dealing with rational inequalities!