### Instruction: Find all solutions to the inequality \[\frac{1}{2} \le \frac{x+1}{x} \le \frac{3}{2}.\].### Response: To solve an inequality involving fractions, I usually try to clear the denominators first by multiplying both sides by a common multiple of them.
In this case, the common denominator is x, so I multiply both sides by x.
However, I have to be careful about the sign of x, because if x is negative, then multiplying by x will reverse the direction of the inequality.
So I have to consider two cases: when x is positive and when x is negative.
If x is positive, then multiplying by x does not change the direction of the inequality, so I get \[ \frac{x}{2} \le x + 1 \le \frac{3x}{2}.\]
To find the solutions in this case, I can split the inequality into two parts and solve them separately: \[ \frac{x}{2} \le x + 1 \quad \text{and} \quad x + 1 \le \frac{3x}{2}.\]
For the first part, I can subtract x from both sides and get \[-\frac{x}{2} \le 1,\] which simplifies to \[x \ge -2.\]
For the second part, I can subtract x from both sides and get \[1 \le \frac{x}{2},\] which simplifies to \[x \ge 2.\]
Combining these two results, I see that x must satisfy both \[x \ge -2 \quad \text{and} \quad x \ge 2,\] which means that x must be greater than or equal to 2.
So the solutions in this case are \[x \in [2, \infty).\]
If x is negative, then multiplying by x will change the direction of the inequality, so I get \[ \frac{x}{2} \ge x + 1 \ge \frac{3x}{2}.\]
To find the solutions in this case, I can again split the inequality into two parts and solve them separately: \[ \frac{x}{2} \ge x + 1 \quad \text{and} \quad x + 1 \ge \frac{3x}{2}.\]
For the first part, I can subtract x from both sides and get \[-\frac{x}{2} \ge 1,\] which simplifies to \[x \le -2.\]
For the second part, I can subtract x from both sides and get \[1 \ge \frac{x}{2},\] which simplifies to \[x \le 2.\]
Combining these two results, I see that x must satisfy both \[x \le -2 \quad \text{and} \quad x \le 2,\] which means that x must be less than or equal to -2.
So the solutions in this case are \[x \in (-\infty, -2].\]
Finally, I can combine the solutions from both cases by taking their union.
The solutions to the original inequality are \[x \in (-\infty, -2] \cup [2, \infty).\]