Solving rational inequalities on a worksheet with handwritten annotations.

Let's solve each of the rational inequalities step by step. We'll go through a, b, c, and d, carefully analyzing the expressions, identifying restrictions (where the denominator is zero), and solving the inequalities using sign analysis.

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a.

$$
\frac{x^2}{x+3} < \frac{9}{x+3}
$$

#### Step 1: Identify restriction
The denominator $ x + 3 $ cannot be zero:
$$
x \ne -3
$$

#### Step 2: Subtract to bring all terms to one side
$$
\frac{x^2}{x+3} - \frac{9}{x+3} < 0
\Rightarrow \frac{x^2 - 9}{x+3} < 0
$$

Factor numerator:
$$
x^2 - 9 = (x - 3)(x + 3)
$$

So:
$$
\frac{(x - 3)(x + 3)}{x + 3} < 0
$$

Now simplify — but only when $ x \ne -3 $ (already restricted):

For $ x \ne -3 $, we can cancel $ x + 3 $, but must consider the sign change depending on whether $ x + 3 > 0 $ or $ < 0 $. However, since we are simplifying a fraction, let’s analyze:

$$
\frac{(x - 3)(x + 3)}{x + 3} = x - 3 \quad \text{for } x \ne -3
$$

But we cannot just cancel unless we consider the sign. So instead, let’s keep it as:

$$
\frac{(x - 3)(x + 3)}{x + 3} < 0
$$

This simplifies to $ x - 3 < 0 $ only if $ x + 3 > 0 $ (i.e., $ x > -3 $), because then the denominator is positive and doesn't affect inequality direction.

But better approach: Use sign chart.

We have:
$$
\frac{(x - 3)(x + 3)}{x + 3} < 0
$$

Note: $ x + 3 $ appears in both numerator and denominator. But we cannot cancel directly because the expression is undefined at $ x = -3 $, and the sign depends on the multiplicity.

Instead, observe that for $ x \ne -3 $, this simplifies to:
$$
x - 3 < 0 \quad \text{if } x + 3 > 0 \quad (\text{i.e., } x > -3)
$$
and
$$
x - 3 > 0 \quad \text{if } x + 3 < 0 \quad (\text{i.e., } x < -3)
$$

Wait — actually, let's do this correctly with sign analysis.

Let’s write:
$$
\frac{(x - 3)(x + 3)}{x + 3} = x - 3 \quad \text{when } x \ne -3
$$

So the expression becomes:
$$
x - 3 < 0 \quad \text{for } x \ne -3
\Rightarrow x < 3 \quad \text{and } x \ne -3
$$

But wait — is that valid? Let's test values.

Try $ x = 0 $:
LHS: $ \frac{0}{3} = 0 $, RHS: $ \frac{9}{3} = 3 $ → $ 0 < 3 $: true ✔

Try $ x = 4 $:
$ \frac{16}{7} \approx 2.29 $, $ \frac{9}{7} \approx 1.29 $ → $ 2.29 < 1.29 $? No ✘

Try $ x = -4 $:
$ \frac{16}{-1} = -16 $, $ \frac{9}{-1} = -9 $ → $ -16 < -9 $? Yes ✔

But $ x = -4 < -3 $

Try $ x = -2 $:
$ \frac{4}{1} = 4 $, $ \frac{9}{1} = 9 $ → $ 4 < 9 $? Yes ✔

So at $ x = -4 $: true
At $ x = -2 $: true
At $ x = 4 $: false

But what about near $ x = -3 $? Try $ x = -3.1 $:
$ x^2 = 9.61 $, $ x + 3 = -0.1 $ → LHS = $ 9.61 / (-0.1) = -96.1 $
RHS = $ 9 / (-0.1) = -90 $ → Is $ -96.1 < -90 $? Yes ✔

Try $ x = -2.9 $:
$ x^2 = 8.41 $, $ x + 3 = 0.1 $ → LHS = 84.1, RHS = 90 → $ 84.1 < 90 $? Yes ✔

So seems like it's true everywhere except possibly around $ x = 3 $

Wait — earlier we saw at $ x = 4 $: $ 16/7 \approx 2.29 $, $ 9/7 \approx 1.29 $, so $ 2.29 < 1.29 $? No.

So inequality fails when $ x > 3 $

But what about $ x < -3 $? It works.

So let's go back to the simplified form:

$$
\frac{x^2 - 9}{x + 3} < 0
\Rightarrow \frac{(x - 3)(x + 3)}{x + 3} < 0
$$

Now, for $ x \ne -3 $, this is equal to $ x - 3 $ only if $ x + 3 > 0 $, but if $ x + 3 < 0 $, then dividing by negative reverses inequality.

But better: factor and use sign chart.

We have:
$$
\frac{(x - 3)(x + 3)}{x + 3}
$$

But notice: $ (x + 3) $ in numerator and denominator. So we can write:

For $ x \ne -3 $, this is:
$$
x - 3 \quad \text{if } x + 3 > 0 \quad (\text{no sign change})
$$
But actually, since $ x + 3 $ cancels algebraically except at $ x = -3 $, and since the function is continuous elsewhere, we can say:

$$
f(x) = \frac{(x - 3)(x + 3)}{x + 3} = x - 3 \quad \text{for } x \ne -3
$$

So the inequality becomes:
$$
x - 3 < 0 \quad \text{and } x \ne -3
\Rightarrow x < 3 \quad \text{and } x \ne -3
$$

So solution:
$$
x < 3, \quad x \ne -3
$$

But let’s verify with $ x = -3 $: undefined → excluded.

So final answer:
$$
(-\infty, -3) \cup (-3, 3)
$$

✔ Answer a: $ \boxed{(-\infty, -3) \cup (-3, 3)} $

---

b.

$$
\frac{4}{x} \geq \frac{3}{x^2}
$$

#### Step 1: Restrictions
Denominators: $ x \ne 0 $

#### Step 2: Bring all terms to one side
$$
\frac{4}{x} - \frac{3}{x^2} \geq 0
$$

Common denominator: $ x^2 $
$$
\frac{4x}{x^2} - \frac{3}{x^2} = \frac{4x - 3}{x^2} \geq 0
$$

So:
$$
\frac{4x - 3}{x^2} \geq 0
$$

Now analyze sign.

Critical points:
- Numerator: $ 4x - 3 = 0 \Rightarrow x = \frac{3}{4} $
- Denominator: $ x^2 = 0 \Rightarrow x = 0 $ (excluded)

Sign chart intervals: $ (-\infty, 0), (0, \frac{3}{4}), (\frac{3}{4}, \infty) $

| Interval | $ 4x - 3 $ | $ x^2 $ | Expression |
|----------------|-------------|----------|------------|
| $ (-\infty, 0) $ | negative | positive | negative |
| $ (0, \frac{3}{4}) $ | negative | positive | negative |
| $ (\frac{3}{4}, \infty) $ | positive | positive | positive |

We want $ \geq 0 $ → includes zero.

Set numerator = 0: $ x = \frac{3}{4} $ → included

Denominator never zero except at $ x = 0 $, which is excluded.

So expression ≥ 0 only when:
- $ x > \frac{3}{4} $
- OR $ x = \frac{3}{4} $

But not at $ x = 0 $

So solution:
$$
\left[\frac{3}{4}, \infty\right)
$$

✔ Answer b: $ \boxed{\left[\frac{3}{4}, \infty\right)} $

---

c.

$$
\frac{5(x - 1)(x + 4)}{x^2(x - 3)} \leq 0
$$

#### Step 1: Restrictions
Denominator: $ x^2(x - 3) \ne 0 \Rightarrow x \ne 0, x \ne 3 $

#### Step 2: Critical points
Numerator zeros: $ x = 1, x = -4 $
Denominator zeros: $ x = 0, x = 3 $

So critical points: $ -4, 0, 1, 3 $

Intervals:
1. $ (-\infty, -4) $
2. $ (-4, 0) $
3. $ (0, 1) $
4. $ (1, 3) $
5. $ (3, \infty) $

We will test sign in each interval.

Expression: $ \frac{5(x - 1)(x + 4)}{x^2(x - 3)} $

Note: $ 5 > 0 $, $ x^2 > 0 $ for $ x \ne 0 $

So sign depends on:
- $ (x - 1) $
- $ (x + 4) $
- $ (x - 3) $

But $ x^2 $ is always positive (except at 0 where undefined), so it doesn’t affect sign.

So effectively, sign of expression is same as sign of:
$$
\frac{(x - 1)(x + 4)}{x - 3}
$$

But we must remember $ x \ne 0, 3 $

Now build sign chart:

| Interval | $ x + 4 $ | $ x - 1 $ | $ x - 3 $ | $ x^2 $ | Overall sign |
|------------------|-----------|-----------|-----------|---------|--------------|
| $ (-\infty, -4) $ | – | – | – | + | (–)(–)/(–) = (–)/– = + → positive |
| $ (-4, 0) $ | + | – | – | + | (+)(–)/(–) = (–)/– = + → positive |
| $ (0, 1) $ | + | – | – | + | (+)(–)/(–) = (–)/– = + → positive |
| $ (1, 3) $ | + | + | – | + | (+)(+)/– = +/– = – → negative |
| $ (3, \infty) $ | + | + | + | + | (+)(+)/+ = + → positive |

Now check equality: $ \leq 0 $ → include where expression = 0

Expression = 0 when numerator = 0: $ x = 1 $ or $ x = -4 $

Check if allowed:
- $ x = 1 $: not in denominator → OK
- $ x = -4 $: OK

So include $ x = -4 $ and $ x = 1 $

But exclude $ x = 0 $ and $ x = 3 $

From sign chart:
- Only negative in $ (1, 3) $
- Zero at $ x = -4 $ and $ x = 1 $

So solution:
- $ x = -4 $
- $ x = 1 $
- $ (1, 3) $

But $ (1, 3) $ is already open at 1, so we combine:

$$
\{-4\} \cup [1, 3)
$$

Wait — at $ x = 1 $: expression = 0 → included

And from $ (1, 3) $: negative → included

So $ [1, 3) $

And $ x = -4 $

So total:
$$
\{-4\} \cup [1, 3)
$$

✔ Answer c: $ \boxed{\{-4\} \cup [1, 3)} $

---

d.

$$
x - \frac{6}{x} \geq 0
$$

#### Step 1: Restriction
$ x \ne 0 $

#### Step 2: Combine terms
$$
\frac{x^2 - 6}{x} \geq 0
$$

So:
$$
\frac{x^2 - 6}{x} \geq 0
$$

Factor numerator: $ x^2 - 6 = (x - \sqrt{6})(x + \sqrt{6}) $

Critical points: $ x = -\sqrt{6}, 0, \sqrt{6} $

Intervals:
1. $ (-\infty, -\sqrt{6}) $
2. $ (-\sqrt{6}, 0) $
3. $ (0, \sqrt{6}) $
4. $ (\sqrt{6}, \infty) $

Sign chart:

| Interval | $ x^2 - 6 $ | $ x $ | Expression |
|--------------------|---------------|--------|------------|
| $ (-\infty, -\sqrt{6}) $ | + | – | + / – = – |
| $ (-\sqrt{6}, 0) $ | – | – | – / – = + |
| $ (0, \sqrt{6}) $ | – | + | – / + = – |
| $ (\sqrt{6}, \infty) $ | + | + | + / + = + |

We want $ \geq 0 $ → include where expression = 0

Set numerator = 0: $ x = \pm\sqrt{6} $

Denominator ≠ 0 → $ x \ne 0 $

So include $ x = -\sqrt{6} $ and $ x = \sqrt{6} $

Now, from sign chart:
- Positive in $ (-\sqrt{6}, 0) $ and $ (\sqrt{6}, \infty) $
- Zero at $ x = \pm\sqrt{6} $

So solution:
- $ [-\sqrt{6}, 0) $: includes $ -\sqrt{6} $, excludes 0
- $ [\sqrt{6}, \infty) $

So:
$$
[-\sqrt{6}, 0) \cup [\sqrt{6}, \infty)
$$

✔ Answer d: $ \boxed{[-\sqrt{6}, 0) \cup [\sqrt{6}, \infty)} $

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✔ Final Answers:

a. $ \boxed{(-\infty, -3) \cup (-3, 3)} $
b. $ \boxed{\left[\frac{3}{4}, \infty\right)} $
c. $ \boxed{\{-4\} \cup [1, 3)} $
d. $ \boxed{[-\sqrt{6}, 0) \cup [\sqrt{6}, \infty)} $

Let me know if you'd like a graph or further explanation!