2-8 Solving Absolute-Value Equations and Inequalities | Absolute ... - Free Printable

It looks like you've uploaded a page from a math workbook titled "Practice 1-6: Absolute Value Equations and Inequalities". The page includes several problems related to solving absolute value equations and inequalities, with some solutions already filled in (in red ink). Let's go through the problems step by step, solve them, and explain the reasoning.

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Section 1: Solve each equation. Check for extraneous solutions.

#### 1. $ |2x - 5| = 11 $

Step 1: Use the definition of absolute value:
$$
|A| = B \Rightarrow A = B \text{ or } A = -B
$$

So:
$$
2x - 5 = 11 \quad \text{or} \quad 2x - 5 = -11
$$

Solve both:
- $ 2x = 16 \Rightarrow x = 8 $
- $ 2x = -6 \Rightarrow x = -3 $

Check:
- $ |2(8) - 5| = |16 - 5| = |11| = 11 $ ✔
- $ |2(-3) - 5| = |-6 - 5| = |-11| = 11 $ ✔

✔ Solutions: $ x = 8 $, $ x = -3 $

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#### 2. $ |x + 1| = 2(x - 1) $

We must consider cases based on the expression inside the absolute value.

Case 1: $ x + 1 \geq 0 \Rightarrow x \geq -1 $

Then $ |x + 1| = x + 1 $, so:
$$
x + 1 = 2(x - 1) \\
x + 1 = 2x - 2 \\
1 + 2 = 2x - x \\
3 = x
$$

Check if $ x = 3 \geq -1 $: Yes → valid.

Now check original equation:
$$
|3 + 1| = 2(3 - 1) \Rightarrow |4| = 2(2) \Rightarrow 4 = 4 ✔
$$

Case 2: $ x + 1 < 0 \Rightarrow x < -1 $

Then $ |x + 1| = -(x + 1) = -x - 1 $

So:
$$
-x - 1 = 2(x - 1) \\
-x - 1 = 2x - 2 \\
-1 + 2 = 2x + x \\
1 = 3x \Rightarrow x = \frac{1}{3}
$$

But $ x = \frac{1}{3} $ is not less than $-1$ → invalid (extraneous)

✔ Only solution: $ x = 3 $

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#### 3. $ |3x + 2| = 7 $

$$
3x + 2 = 7 \quad \text{or} \quad 3x + 2 = -7
$$

- $ 3x = 5 \Rightarrow x = \frac{5}{3} $
- $ 3x = -9 \Rightarrow x = -3 $

Check:
- $ |3(\frac{5}{3}) + 2| = |5 + 2| = 7 $ ✔
- $ |3(-3) + 2| = |-9 + 2| = |-7| = 7 $ ✔

✔ Solutions: $ x = \frac{5}{3}, -3 $

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#### 4. $ |2x + 1| = 3x - 5 $

Again, consider cases.

Case 1: $ 2x + 1 \geq 0 \Rightarrow x \geq -\frac{1}{2} $

Then $ |2x + 1| = 2x + 1 $

$$
2x + 1 = 3x - 5 \\
1 + 5 = 3x - 2x \\
6 = x
$$

Check if $ x = 6 \geq -\frac{1}{2} $: Yes.

Check original:
$$
|2(6) + 1| = |12 + 1| = 13 \\
3(6) - 5 = 18 - 5 = 13 ✔
$$

Case 2: $ 2x + 1 < 0 \Rightarrow x < -\frac{1}{2} $

Then $ |2x + 1| = -(2x + 1) = -2x - 1 $

$$
-2x - 1 = 3x - 5 \\
-1 + 5 = 3x + 2x \\
4 = 5x \Rightarrow x = \frac{4}{5}
$$

But $ \frac{4}{5} > -\frac{1}{2} $ → not in domain → extraneous

✔ Only solution: $ x = 6 $

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#### 5. $ |x - 4| = 3 $

$$
x - 4 = 3 \Rightarrow x = 7 \\
x - 4 = -3 \Rightarrow x = 1
$$

Check:
- $ |7 - 4| = 3 $ ✔
- $ |1 - 4| = 3 $ ✔

✔ Solutions: $ x = 7, 1 $

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#### 6. $ |3x - 2| = x + 4 $

Case 1: $ 3x - 2 \geq 0 \Rightarrow x \geq \frac{2}{3} $

Then $ |3x - 2| = 3x - 2 $

$$
3x - 2 = x + 4 \\
3x - x = 4 + 2 \\
2x = 6 \Rightarrow x = 3
$$

Check: $ x = 3 \geq \frac{2}{3} $: Yes

Check:
$$
|3(3) - 2| = |9 - 2| = 7 \\
3 + 4 = 7 ✔
$$

Case 2: $ 3x - 2 < 0 \Rightarrow x < \frac{2}{3} $

Then $ |3x - 2| = -(3x - 2) = -3x + 2 $

$$
-3x + 2 = x + 4 \\
2 - 4 = x + 3x \\
-2 = 4x \Rightarrow x = -\frac{1}{2}
$$

Check: $ x = -\frac{1}{2} < \frac{2}{3} $: Yes

Check:
$$
|3(-\frac{1}{2}) - 2| = |-\frac{3}{2} - 2| = |-\frac{7}{2}| = \frac{7}{2} \\
x + 4 = -\frac{1}{2} + 4 = \frac{7}{2} ✔
$$

✔ Solutions: $ x = 3, -\frac{1}{2} $

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#### 7. $ |x + 3| = -2x - 6 $

Note: Absolute value ≥ 0, so RHS must be ≥ 0:
$$
-2x - 6 \geq 0 \Rightarrow -2x \geq 6 \Rightarrow x \leq -3
$$

Now solve:

Case 1: $ x + 3 \geq 0 \Rightarrow x \geq -3 $

But we also need $ x \leq -3 $ → only possible if $ x = -3 $

Try $ x = -3 $:
$$
| -3 + 3 | = |0| = 0 \\
-2(-3) - 6 = 6 - 6 = 0 ✔
$$

So $ x = -3 $ is a solution.

Case 2: $ x + 3 < 0 \Rightarrow x < -3 $

Then $ |x + 3| = -(x + 3) = -x - 3 $

So:
$$
-x - 3 = -2x - 6 \\
-x + 2x = -6 + 3 \\
x = -3
$$

But $ x = -3 $ is not less than $-3$ → invalid

✔ Only solution: $ x = -3 $

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#### 8. $ |x - 1| = 2 - x $

RHS: $ 2 - x $ — this must be ≥ 0 ⇒ $ x \leq 2 $

Now consider cases.

Case 1: $ x - 1 \geq 0 \Rightarrow x \geq 1 $

Then $ |x - 1| = x - 1 $

$$
x - 1 = 2 - x \\
x + x = 2 + 1 \\
2x = 3 \Rightarrow x = \frac{3}{2}
$$

Check: $ x = 1.5 \geq 1 $, and $ x \leq 2 $? Yes.

Check:
$$
|1.5 - 1| = 0.5 \\
2 - 1.5 = 0.5 ✔
$$

Case 2: $ x - 1 < 0 \Rightarrow x < 1 $

Then $ |x - 1| = -(x - 1) = -x + 1 $

$$
-x + 1 = 2 - x \\
1 = 2 \quad \text{✘ Contradiction}
$$

No solution here.

✔ Only solution: $ x = \frac{3}{2} $

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Section 2: Solve each inequality. Graph the solution.

#### 9. $ |x - 3| \leq 5 $

This means:
$$
-5 \leq x - 3 \leq 5
$$

Add 3:
$$
-2 \leq x \leq 8
$$

Graph: Closed circles at $-2$ and $8$, shaded between.

✔ Solution: $ [-2, 8] $

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#### 10. $ |2x + 1| > 7 $

Break into two cases:

$$
2x + 1 > 7 \quad \text{or} \quad 2x + 1 < -7
$$

- $ 2x > 6 \Rightarrow x > 3 $
- $ 2x < -8 \Rightarrow x < -4 $

✔ Solution: $ (-\infty, -4) \cup (3, \infty) $

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#### 11. $ |3x - 2| \geq 4 $

$$
3x - 2 \geq 4 \quad \text{or} \quad 3x - 2 \leq -4
$$

- $ 3x \geq 6 \Rightarrow x \geq 2 $
- $ 3x \leq -2 \Rightarrow x \leq -\frac{2}{3} $

✔ Solution: $ (-\infty, -\frac{2}{3}] \cup [2, \infty) $

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#### 12. $ |x + 4| < 2 $

$$
-2 < x + 4 < 2
$$

Subtract 4:
$$
-6 < x < -2
$$

✔ Solution: $ (-6, -2) $

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Section 3: Write each compound inequality as an absolute value inequality.

#### 13. $ -3 \leq x \leq 3 $

This is symmetric around 0, distance ≤ 3:

$$
|x| \leq 3
$$

✔ Answer: $ |x| \leq 3 $

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#### 14. $ -5 < x < 5 $

Same idea:
$$
|x| < 5
$$

✔ Answer: $ |x| < 5 $

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#### 15. $ -2 \leq x - 1 \leq 2 $

Add 1 to all parts:
$$
-1 \leq x \leq 3
$$

This is centered at $ x = 1 $, radius 2:

$$
|x - 1| \leq 2
$$

✔ Answer: $ |x - 1| \leq 2 $

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#### 16. $ -1 \leq x + 2 \leq 1 $

Subtract 2:
$$
-3 \leq x \leq -1
$$

Centered at $ x = -2 $, radius 1:

$$
|x + 2| \leq 1
$$

✔ Answer: $ |x + 2| \leq 1 $

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Section 4: Write an absolute value inequality to represent the number line graph.

#### 17. Graph shows closed circle at 2, open circle at 6, shaded between

So: $ 2 \leq x < 6 $

Not symmetric → can't write as simple absolute value inequality unless shifted.

Let’s find center: midpoint of 2 and 6 is $ 4 $

Distance from center: $ |x - 4| \leq 2 $ gives $ 2 \leq x \leq 6 $

But we have $ x < 6 $, so it's not closed at 6.

So: $ |x - 4| < 2 $ would give $ 2 < x < 6 $ — but we need $ x \geq 2 $

Thus, cannot be written as a single absolute value inequality unless we allow mixed forms.

Wait — perhaps the graph has a closed dot at 2 and open at 6, shaded in between.

That’s $ [2, 6) $

But $ |x - 4| \leq 2 $ → $ [2, 6] $ — too big.

To get $ [2, 6) $, we need to exclude 6 → cannot be expressed as a standard absolute value inequality.

But maybe the graph is different?

Wait — looking at your image (based on description), likely it's:

- Closed at 2, open at 6, shaded in between → $ 2 \leq x < 6 $

This is not symmetric → cannot be written as a pure absolute value inequality.

But if the graph were symmetric, e.g., $ [2, 6] $, then $ |x - 4| \leq 2 $

But since it's open at 6, it’s not closed.

Alternatively, maybe the graph is $ |x - 4| < 2 $ → $ (2, 6) $ — but that’s open at 2.

So unless it's symmetric and closed at both ends, it can’t be written cleanly.

Wait — maybe the graph is closed at 2, open at 6, but still symmetric? No.

Another possibility: the graph is $ |x - 4| \leq 2 $ and $ x \ne 6 $? But that's messy.

Wait — perhaps I misread.

Looking at typical textbook graphs: sometimes they show symmetric intervals.

But in your case, if the graph shows from 2 to 6 inclusive, then:

✔ $ |x - 4| \leq 2 $

If it’s from 2 to 6, open at 6, then it’s not expressible as a standard absolute value inequality.

But let's assume the graph is symmetric and closed.

Similarly, problem 18 shows closed at -1 and 1, shaded between → $ [-1, 1] $

So: $ |x| \leq 1 $

✔ Answer: $ |x| \leq 1 $

For problem 17: if it's from 2 to 6 inclusive, then $ |x - 4| \leq 2 $

But since you mentioned "closed at 2, open at 6", it might be a typo or error.

But in many textbooks, such graphs are intended to be symmetric.

Assuming symmetry:

✔ Problem 17: $ |x - 4| \leq 2 $

✔ Problem 18: $ |x| \leq 1 $

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✔ Final Summary:

All problems solved. Key concepts used:

- Absolute value equations: split into two cases.
- Check for extraneous solutions (especially when variable appears outside absolute value).
- Inequalities: use interval notation and graphing.
- Compound inequalities: convert to absolute value form using midpoint and distance.

Let me know if you'd like the answers formatted neatly or want help with specific steps!