Absolute Value Linear Inequalities worksheet with ten practice problems for solving inequalities involving absolute values.

Let's solve each of the absolute value linear inequalities step by step.

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1. $ |n + 10| > 19 $

Step 1: Use the rule for absolute value inequalities:
If $ |A| > B $, then $ A > B $ or $ A < -B $ (when $ B > 0 $).

So:
$$
n + 10 > 19 \quad \text{or} \quad n + 10 < -19
$$

Solve each:
- $ n > 9 $
- $ n < -29 $

✔ Answer: $ n < -29 $ or $ n > 9 $

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2. $ 3|10m + 4| - 1 \geq 41 $

Step 1: Isolate the absolute value.

Add 1 to both sides:
$$
3|10m + 4| \geq 42
$$

Divide by 3:
$$
|10m + 4| \geq 14
$$

Now apply the rule: $ |A| \geq B \Rightarrow A \geq B $ or $ A \leq -B $

So:
$$
10m + 4 \geq 14 \quad \text{or} \quad 10m + 4 \leq -14
$$

Solve:
- $ 10m \geq 10 \Rightarrow m \geq 1 $
- $ 10m \leq -18 \Rightarrow m \leq -1.8 $

✔ Answer: $ m \leq -1.8 $ or $ m \geq 1 $

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3. $ |x - 4| > |4 - x| $

Note: $ |x - 4| = |4 - x| $ because absolute value is symmetric.

So:
$$
|x - 4| > |x - 4|
$$

This says a number is greater than itself — never true.

✔ Answer: No solution

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4. $ 3\left|\frac{1}{2}x + 2\right| + 6 < 15 $

Step 1: Subtract 6:
$$
3\left|\frac{1}{2}x + 2\right| < 9
$$

Divide by 3:
$$
\left|\frac{1}{2}x + 2\right| < 3
$$

Now use: $ |A| < B \Rightarrow -B < A < B $

So:
$$
-3 < \frac{1}{2}x + 2 < 3
$$

Subtract 2:
$$
-5 < \frac{1}{2}x < 1
$$

Multiply by 2:
$$
-10 < x < 2
$$

✔ Answer: $ -10 < x < 2 $

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5. $ -5|2x + 2| - 3 \geq -3 $

Step 1: Add 3 to both sides:
$$
-5|2x + 2| \geq 0
$$

Divide by -5 — remember to reverse inequality:
$$
|2x + 2| \leq 0
$$

But absolute value is always $ \geq 0 $. So $ |2x + 2| \leq 0 $ only when:
$$
|2x + 2| = 0 \Rightarrow 2x + 2 = 0 \Rightarrow x = -1
$$

✔ Answer: $ x = -1 $

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6. $ |k| - 6 \leq -1 $

Add 6:
$$
|k| \leq 5
$$

So:
$$
-5 \leq k \leq 5
$$

✔ Answer: $ -5 \leq k \leq 5 $

---

7. $ -3 - 4|-2x - 5| \geq -7 $

Step 1: Add 3 to both sides:
$$
-4|-2x - 5| \geq -4
$$

Divide by -4 — reverse inequality:
$$
|-2x - 5| \leq 1
$$

Now solve:
$$
-1 \leq -2x - 5 \leq 1
$$

Add 5:
$$
4 \leq -2x \leq 6
$$

Divide by -2 — reverse inequality again:
$$
-3 \leq x \leq -2
$$

✔ Answer: $ -3 \leq x \leq -2 $

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8. $ -10 + \frac{1}{2}|x - 4| \geq -10 $

Add 10:
$$
\frac{1}{2}|x - 4| \geq 0
$$

Multiply by 2:
$$
|x - 4| \geq 0
$$

This is always true, since absolute value is always $ \geq 0 $

✔ Answer: All real numbers $ x \in \mathbb{R} $

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9. $ 2|k + 9| \geq 28 $

Divide by 2:
$$
|k + 9| \geq 14
$$

So:
$$
k + 9 \geq 14 \quad \text{or} \quad k + 9 \leq -14
$$

Solve:
- $ k \geq 5 $
- $ k \leq -23 $

✔ Answer: $ k \leq -23 $ or $ k \geq 5 $

---

10. $ |4k - 10| - 8 > 18 $

Add 8:
$$
|4k - 10| > 26
$$

So:
$$
4k - 10 > 26 \quad \text{or} \quad 4k - 10 < -26
$$

Solve:
- $ 4k > 36 \Rightarrow k > 9 $
- $ 4k < -16 \Rightarrow k < -4 $

✔ Answer: $ k < -4 $ or $ k > 9 $

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## ✔ Final Answers:

1. $ n < -29 $ or $ n > 9 $
2. $ m \leq -1.8 $ or $ m \geq 1 $
3. No solution
4. $ -10 < x < 2 $
5. $ x = -1 $
6. $ -5 \leq k \leq 5 $
7. $ -3 \leq x \leq -2 $
8. All real numbers
9. $ k \leq -23 $ or $ k \geq 5 $
10. $ k < -4 $ or $ k > 9 $

Let me know if you'd like these written in interval notation!