How to find the smallest positive integer of the inequality |4x+2|<|6x-8|

To solve the inequality \( |4x + 2| < |6x - 8| \), we need to break it down by considering different cases based on when the absolute values change their behavior. Absolute value inequalities require analyzing different conditions based on the expressions inside the absolute values.

### Step 1: Define the critical points

First, find where the expressions inside the absolute values are zero, as this will define the boundaries for our cases:

1. \( 4x + 2 = 0 \)  
   \[ x = -\frac{1}{2} \]

2. \( 6x - 8 = 0 \)  
   \[ x = \frac{4}{3} \]

So, the critical points are \( x = -\frac{1}{2} \) and \( x = \frac{4}{3} \). These will divide the number line into three intervals:  
1. \( x < -\frac{1}{2} \)
2. \( -\frac{1}{2} \leq x \leq \frac{4}{3} \)
3. \( x > \frac{4}{3} \)

Now, let's solve the inequality for each interval.

---

### Step 2: Solve for each interval

#### Case 1: \( x < -\frac{1}{2} \)

In this region, both \( 4x + 2 \) and \( 6x - 8 \) are negative. Thus, we can drop the absolute values by reversing their signs:
\[
-(4x + 2) < -(6x - 8)
\]
Simplifying:
\[
-4x - 2 < -6x + 8
\]
Add \( 6x \) to both sides:
\[
2x - 2 < 8
\]
Add 2 to both sides:
\[
2x < 10
\]
Divide by 2:
\[
x < 5
\]
In this case, we are already in the region where \( x < -\frac{1}{2} \), so we combine this with the result to get:
\[
x < -\frac{1}{2}
\]
Thus, for this case, the solution is \( x < -\frac{1}{2} \).

---

#### Case 2: \( -\frac{1}{2} \leq x \leq \frac{4}{3} \)

In this interval, \( 4x + 2 \) is non-negative (positive or zero), but \( 6x - 8 \) is negative. Thus, we can drop the absolute values as follows:
\[
4x + 2 < -(6x - 8)
\]
Simplify:
\[
4x + 2 < -6x + 8
\]
Add \( 6x \) to both sides:
\[
10x + 2 < 8
\]
Subtract 2 from both sides:
\[
10x < 6
\]
Divide by 10:
\[
x < \frac{3}{5}
\]
Thus, for this case, the solution is \( -\frac{1}{2} \leq x < \frac{3}{5} \).

---

#### Case 3: \( x > \frac{4}{3} \)

In this region, both \( 4x + 2 \) and \( 6x - 8 \) are positive, so we can drop the absolute values directly:
\[
4x + 2 < 6x - 8
\]
Simplifying:
\[
4x + 2 < 6x - 8
\]
Subtract \( 4x \) from both sides:
\[
2 < 2x - 8
\]
Add 8 to both sides:
\[
10 < 2x
\]
Divide by 2:
\[
5 < x
\]
Thus, for this case, the solution is \( x > 5 \).

---

### Step 3: Combine the results

- Case 1 gave us \( x < -\frac{1}{2} \).
- Case 2 gave us \( -\frac{1}{2} \leq x < \frac{3}{5} \).
- Case 3 gave us \( x > 5 \).

Thus, the solution to the inequality \( |4x + 2| < |6x - 8| \) is:
\[
x < \frac{3}{5} \quad \text{or} \quad x > 5
\]

### Step 4: Find the smallest positive integer

The smallest positive integer that satisfies this inequality is \( x = 6 \), since the second condition is \( x > 5 \).

---

Let me know if you'd like further details or explanations on any step! 

Here are 5 related questions to explore:
1. What happens if the inequality was \( |4x + 2| > |6x - 8| \)?
2. How do you solve a system of absolute value inequalities?
3. What are the general rules for solving absolute value equations?
4. How do inequalities change if we have square roots instead of absolute values?
5. Can absolute value inequalities have no solution? Why?

**Tip:** Always check the critical points when working with absolute value inequalities to ensure you don’t miss any cases!

Learn how to solve the inequality |4x+2| < |6x-8| with a step-by-step solution. This problem covers key algebraic concepts such as handling absolute value inequalities and finding critical points. Suitable for students in Grades 10-12.

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