Solve the inequality (x^2 - 4x + 3)/(x^2 - 9) <= 0.

Step-by-step solution

1. 1

   Factor the numerator and the denominator.

   $x^2-4x+3=(x-1)(x-3),\qquad x^2-9=(x-3)(x+3)$

   so the inequality is

   $\frac{(x-1)(x-3)}{(x-3)(x+3)}\le0$

2. 2

   Record the domain before cancelling anything. The denominator vanishes at $x=3$ and $x=-3$, so both are excluded from the start:

   $x\neq3,\qquad x\neq-3$

   This must be written down first, because the next step makes $x=3$ invisible.

3. 3

   Cancel the common factor, keeping the exclusion. For $x\neq3$ the shared $(x-3)$ divides out:

   $\frac{x-1}{x+3}\le0\qquad(x\neq3,\ x\neq-3)$

   The cancellation creates a removable hole at $x=3$, not a sign change — the original function simply has no value there, while the simplified one does. Forgetting this is the standard trap.

4. 4

   Find the critical points of the reduced fraction and test the intervals. They are $x=1$ (numerator zero) and $x=-3$ (denominator zero). At $x=-4$: $\tfrac{-5}{-1}=5>0$. At $x=0$: $\tfrac{-1}{3}<0$. At $x=2$: $\tfrac{1}{5}>0$. So only $(-3,1)$ is negative.

5. 5

   Decide the endpoints. At $x=1$ the fraction is $0$, and $\le0$ includes zero, so $x=1$ is in. At $x=-3$ the expression is undefined, so it is out. And $x=3$, though it sits outside $(-3,1]$ anyway, remains excluded by the domain — worth stating so the reasoning is complete.

6. 6

   State the answer and verify.

   $x\in(-3,1]$

   Evaluating the original unsimplified fraction with exact fractions at $4001$ rational points on $[-10,10]$ (skipping $x=\pm3$), the inequality holds at precisely the points of $(-3,1]$ ✓.