To solve this inequality, I want to isolate $x$ on one side of the inequality sign. But before I do that, I need to be careful about the denominator, $x+3$, since it cannot be zero. So I will exclude $x = -3$ from the possible solutions, since that would make the fraction undefined. Now, I can multiply both sides of the inequality by $x+3$, but I need to consider the sign of $x+3$. If $x+3 > 0$, then multiplying by it does not change the inequality sign. If $x+3 < 0$, then multiplying by it reverses the inequality sign. So I need to split the inequality into two cases: one where $x+3 > 0$ and one where $x+3 < 0$. In the first case, where $x+3 > 0$, I can multiply both sides by $x+3$ and get \[2x-5 \ge 2(x+3).\] Expanding the right side, I get \[2x-5 \ge 2x+6.\] Subtracting $2x$ from both sides, I get \[-5 \ge 6.\] This is a false statement, so there are no solutions for $x$ when $x+3 > 0$. In the second case, where $x+3 < 0$, I can multiply both sides by $x+3$ and get \[2x-5 \le 2(x+3).\] Expanding the right side, I get \[2x-5 \le 2x+6.\] Subtracting $2x$ from both sides, I get \[-5 \le 6.\] This is a true statement, so there are solutions for $x$ when $x+3 < 0$. To find those solutions, I need to solve the inequality $x+3 < 0$. Subtracting $3$ from both sides, I get \[x < -3.\] This means that any value of $x$ that is less than $-3$ is a solution to the original inequality. To write the solution in interval notation, I use parentheses to indicate that the endpoints are not included, and I use infinity to indicate that there is no upper bound. So the final answer is \[(-\infty, -3).\] 
Find all real numbers $x$ satisfying \[\frac{2x-5}{x+3} \ge 2.\](Give your answer in interval notation.)