### Instruction: Find all solutions to the inequality \[\frac{1}{2} \le \frac{x+1}{x} \le \frac{3}{2}.\].### Response: I want to solve for x, so I need to get rid of the fractions.
To do that, I can multiply both sides of the inequality by x, but I have to be careful about the sign of x.
If x is positive, then the inequality stays the same, but if x is negative, then the inequality flips.
So I have to consider both cases separately.
Case 1: x > 0.
Then I can multiply by x and get \[x \le x+1 \le \frac{3}{2}x.\]
Subtracting x from all parts, I get \[0 \le 1 \le \frac{1}{2}x.\]
Dividing by 1/2, I get \[0 \le 2 \le x.\]
This means that x is any positive number greater than or equal to 2.
Case 2: x < 0.
Then I have to multiply by x and flip the inequality, so I get \[x \ge x+1 \ge \frac{3}{2}x.\]
Subtracting x from all parts, I get \[0 \ge 1 \ge \frac{1}{2}x.\]
Dividing by 1/2, I get \[0 \ge 2 \ge x.\]
This means that x is any negative number less than or equal to -2.
So the final answer is that x is either less than or equal to -2, or greater than or equal to 2.