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Let's go through the inequalities one by one and solve them step-by-step.

### 1. \( 2x - 7 < 4x - 2 \)

**Step 1**: Rearrange the terms to collect like terms:
\[
2x - 7 < 4x - 2
\]
Subtract \(2x\) from both sides:
\[
-7 < 2x - 2
\]
Add 2 to both sides:
\[
-5 < 2x
\]
Now divide by 2:
\[
x > -\frac{5}{2}
\]

So, the solution for the first inequality is \( x > -\frac{5}{2} \).

---

### 2. \( x^2 - x < 6 \)

**Step 1**: Rearrange to set the inequality to zero:
\[
x^2 - x - 6 < 0
\]
Factor the quadratic:
\[
(x - 3)(x + 2) < 0
\]
Now find the critical points, \( x = 3 \) and \( x = -2 \).

The solution comes from the intervals where the product is negative:
- For \( x < -2 \), both terms are negative, so the product is positive.
- For \( -2 < x < 3 \), one term is positive and one term is negative, so the product is negative.
- For \( x > 3 \), both terms are positive, so the product is positive.

So, the solution is \( -2 < x < 3 \).

---

### 3. \( \frac{2}{x} \geq x + 1 \)

**Step 1**: Multiply both sides by \( x \) (keeping in mind that \( x \neq 0 \) and considering the sign of \( x \)):
\[
2 \geq x(x + 1)
\]
This becomes:
\[
2 \geq x^2 + x
\]
Rearrange it:
\[
x^2 + x - 2 \leq 0
\]
Factor the quadratic:
\[
(x - 1)(x + 2) \leq 0
\]
Now, the critical points are \( x = 1 \) and \( x = -2 \).

Using a sign analysis, we find that the product is negative or zero in the interval \( -2 \leq x \leq 1 \).

So, the solution is \( -2 \leq x \leq 1 \).

---

### 4. \( (x + 1)(x - 1)^2(x - 3) \leq 0 \)

The critical points are \( x = -1 \), \( x = 1 \), and \( x = 3 \).

We analyze the sign of the expression in each interval defined by these points:
- For \( x < -1 \), all factors are negative or positive, making the product positive.
- For \( -1 < x < 1 \), the term \( (x + 1) \) is positive, but \( (x - 1)^2 \) is always positive (since it's squared), and \( (x - 3) \) is negative, making the product negative.
- For \( 1 < x < 3 \), both \( (x - 1)^2 \) and \( (x + 1) \) are positive, while \( (x - 3) \) is negative, making the product negative.
- For \( x > 3 \), all terms are positive, making the product positive.

Including the equalities at the critical points, the solution is:
\[
x \in [-1, 3]
\]

---

### 5. \( \frac{x - 1}{x + 2} \geq 0 \)

The critical points are \( x = 1 \) and \( x = -2 \) (though \( x = -2 \) causes a division by zero, so it's excluded).

Now analyze the sign of \( \frac{x - 1}{x + 2} \):
- For \( x < -2 \), both \( x - 1 \) and \( x + 2 \) are negative, so the ratio is positive.
- For \( -2 < x < 1 \), \( x - 1 \) is negative and \( x + 2 \) is positive, making the ratio negative.
- For \( x > 1 \), both \( x - 1 \) and \( x + 2 \) are positive, so the ratio is positive.

So, the solution is \( (-\infty, -2) \cup [1, \infty) \).

---

Would you like further clarifications on any of these steps or additional details?

Here are 5 related questions to deepen your understanding:
1. How do you solve inequalities involving absolute values?
2. What is the geometric interpretation of solving quadratic inequalities?
3. How do you solve compound inequalities?
4. What role do critical points play in solving polynomial inequalities?
5. How do you handle rational inequalities where the denominator can be zero?

**Tip**: Always be careful when multiplying both sides of an inequality by a variable, as it might change the direction of the inequality if the variable is negative.

Solve the following inequalities: 1. 2x - 7 < 4x - 2 2. x^2 - x < 6 3. 2/x ≥ x + 1 4. (x + 1)(x - 1)^2(x - 3) ≤ 0 5. (x - 1)/(x + 2) ≥ 0

Learn how to solve a range of inequalities, from linear and quadratic to rational, with detailed step-by-step explanations. This lesson is ideal for high school students tackling concepts such as critical points, sign analysis, and polynomial factoring.

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