Solve the inequality $|\frac{-6x}{x^2+9}| > 1$.

1. Problem Description

Solve the inequality $|\frac{-6x}{x^2+9}| > 1$.

2. Solution Steps

The given inequality is $|\frac{-6x}{x^2+9}| > 1$.

Since absolute value is always non-negative, we have

$|\frac{-6x}{x^2+9}| = \frac{|-6x|}{|x^2+9|} = \frac{6|x|}{x^2+9}$.

So the inequality becomes $\frac{6|x|}{x^2+9} > 1$.

Since $x^2+9$ is always positive, we can multiply both sides by $x^2+9$ without changing the inequality sign:

$6|x| > x^2+9$.

This can be written as

$x^2 - 6|x| + 9 < 0$.

Let $y = |x|$. Then the inequality becomes

$y^2 - 6y + 9 < 0$.

Factoring the quadratic gives

$(y-3)^2 < 0$.

However, the square of a real number cannot be negative. Therefore, there is no real solution to $(y-3)^2 < 0$.

3. Final Answer

No solution.