### Instruction: Find all real numbers $x$ which satisfy \[\frac{x-1}{x-3} \ge 2.\](Give your answer in interval notation.).### Response: To find the real numbers $x$ that satisfy the 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 $x-3$, because it cannot be zero, otherwise the fraction would be undefined.
So I will first exclude the value $x=3$ from the solution set, since it would make the denominator zero.
Now I can multiply both sides of the inequality by $x-3$, but I have to remember that the inequality sign might change direction depending on the sign of $x-3$.
If $x-3$ is positive, then the inequality sign stays the same, and I get \[x-1 \ge 2(x-3).\]
If $x-3$ is negative, then the inequality sign reverses, and I get \[x-1 \le 2(x-3).\]
I will consider both cases separately.
In the first case, where $x-3$ is positive, I can simplify the inequality by expanding the right side and moving everything to one side, getting \[x-1 - 2x + 6 \ge 0.\]
This simplifies to \[-x+5 \ge 0.\]
To isolate $x$, I can divide both sides by -1, but I have to remember to reverse the inequality sign, getting \[x-5 \le 0.\]
This means that $x$ is less than or equal to 5, but I also have to remember that $x-3$ is positive, which means that $x$ is greater than 3.
So the solution in this case is the interval \[(3,5].\]
In the second case, where $x-3$ is negative, I can simplify the inequality by expanding the right side and moving everything to one side, getting \[x-1 - 2x + 6 \le 0.\]
This simplifies to \[-x+5 \le 0.\]
To isolate $x$, I can divide both sides by -1, but I have to remember to reverse the inequality sign, getting \[x-5 \ge 0.\]
This means that $x$ is greater than or equal to 5, but I also have to remember that $x-3$ is negative, which means that $x$ is less than 3.
So the solution in this case is the empty set, since there are no values of $x$ that satisfy both conditions.