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$$x-2 < \frac{6x-9}{x+2}$$ $$x-2 - \frac{6x-9}{x+2} < 0$$

The common denominator is $(x+2)$.

$$\frac{(x-2)(x+2)}{x+2} - \frac{6x-9}{x+2} < 0$$

Expand the numerator:

$$\frac{x^2 - 4 - (6x-9)}{x+2} < 0$$

Distribute the negative sign:

$$\frac{x^2 - 4 - 6x + 9}{x+2} < 0$$

Combine like terms in the numerator:

$$\frac{x^2 - 6x + 5}{x+2} < 0$$

The numerator $x^2 - 6x + 5$ can be factored as $(x-1)(x-5)$.

$$\frac{(x-1)(x-5)}{x+2} < 0$$

These points divide the number line into intervals.

• From the numerator: $x-1=0 \implies x=1$ and $x-5=0 \implies x=5$.
• From the denominator: $x+2=0 \implies x=-2$. The critical points are $-2, 1, 5$.

• Interval 1: $x < -2$ (e.g., test $x=-3$)

$$\frac{(-3-1)(-3-5)}{-3+2} = \frac{(-4)(-8)}{-1} = \frac{32}{-1} = -32$$

Since $-32 < 0$, this interval $(-\infty, -2)$ is part of the solution.

• Interval 2: $-2 < x < 1$ (e.g., test $x=0$)

$$\frac{(0-1)(0-5)}{0+2} = \frac{(-1)(-5)}{2} = \frac{5}{2}$$

Since $\frac{5}{2} > 0$, this interval is not part of the solution.

• Interval 3: $1 < x < 5$ (e.g., test $x=2$)

$$\frac{(2-1)(2-5)}{2+2} = \frac{(1)(-3)}{4} = \frac{-3}{4}$$

Since $\frac{-3}{4} < 0$, this interval $(1, 5)$ is part of the solution.

• Interval 4: $x > 5$ (e.g., test $x=6$)

$$\frac{(6-1)(6-5)}{6+2} = \frac{(5)(1)}{8} = \frac{5}{8}$$

Since $\frac{5}{8} > 0$, this interval is not part of the solution.

The inequality $\frac{(x-1)(x-5)}{x+2} < 0$ is satisfied when the expression is negative.
Combining the intervals where the expression is negative gives the solution.

The range of values for $x$ is:

$$(-\infty, -2) \cup (1, 5)$$