### Instruction: Find all real numbers $x$ such that \[\left(\frac{x}{2x-1}\right)^2 \le 9.\].### Response: I want to solve this inequality for $x$, so I need to isolate $x$ on one side of the inequality.
To do that, I need to get rid of the square on the left side, so I can take the square root of both sides.
But I have to be careful, because when I take the square root of a positive number, I get two possible values: a positive and a negative one.
So I have to write \[\frac{x}{2x-1} \le \sqrt{9} \quad \text{or} \quad \frac{x}{2x-1} \ge -\sqrt{9}.\]
Now I can simplify the right sides and get \[\frac{x}{2x-1} \le 3 \quad \text{or} \quad \frac{x}{2x-1} \ge -3.\]
To solve these inequalities for $x$, I need to clear the fractions by multiplying both sides by the denominator $2x-1$.
But I have to be careful again, because if the denominator is negative, then the inequality sign will flip.
So I have to consider two cases: when $2x-1 > 0$ and when $2x-1 < 0$.
If $2x-1 > 0$, then I can multiply both sides of the first inequality by $2x-1$ without changing the sign, and I get \[x \le 3(2x-1) \implies x \le 6x-3 \implies -5x \le -3 \implies x \ge \frac{3}{5}.\]
Similarly, I can multiply both sides of the second inequality by $2x-1$ without changing the sign, and I get \[x \ge -3(2x-1) \implies x \ge -6x+3 \implies 7x \ge 3 \implies x \ge \frac{3}{7}.\]
So, if $2x-1 > 0$, then the solution is $x \ge \frac{3}{5}$, since this is the stronger condition than $x \ge \frac{3}{7}$.
If $2x-1 < 0$, then I can multiply both sides of the first inequality by $2x-1$, but I have to flip the sign, and I get \[x \ge 3(2x-1) \implies x \ge 6x-3 \implies -5x \ge -3 \implies x \le \frac{3}{5}.\]
Similarly, I can multiply both sides of the second inequality by $2x-1$, but I have to flip the sign, and I get \[x \le -3(2x-1) \implies x \le -6x+3 \implies 7x \le 3 \implies x \le \frac{3}{7}.\]
So, if $2x-1 < 0$, then the solution is $x \le \frac{3}{7}$, since this is the stronger condition than $x \le \frac{3}{5}$.
Therefore, the final solution is the union of these two cases, which is \[x \le \frac{3}{7} \quad \text{or} \quad x \ge \frac{3}{5}.\]