Absolute Value Inequalities Worksheets with Answer Key - Free Printable

Let's solve each of the absolute value inequalities step by step and express the solution in interval notation. We'll go through each problem carefully.

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1. $ |3x - 4| \leq |x - 3| $

We are solving an inequality with absolute values on both sides.

#### Step 1: Consider cases based on critical points
The expressions inside the absolute values change sign at:
- $ 3x - 4 = 0 \Rightarrow x = \frac{4}{3} $
- $ x - 3 = 0 \Rightarrow x = 3 $

So, we divide the number line into intervals:
- $ (-\infty, \frac{4}{3}) $
- $ [\frac{4}{3}, 3) $
- $ [3, \infty) $

We will analyze each interval.

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#### Case 1: $ x < \frac{4}{3} $

Then:
- $ 3x - 4 < 0 \Rightarrow |3x - 4| = -(3x - 4) = -3x + 4 $
- $ x - 3 < 0 \Rightarrow |x - 3| = -(x - 3) = -x + 3 $

Inequality becomes:
$$
-3x + 4 \leq -x + 3 \\
-3x + 4 \leq -x + 3 \\
-2x \leq -1 \\
x \geq \frac{1}{2}
$$

But this is under the assumption $ x < \frac{4}{3} $. So intersection:
$$
x \in \left[\frac{1}{2}, \frac{4}{3}\right)
$$

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#### Case 2: $ \frac{4}{3} \leq x < 3 $

Then:
- $ 3x - 4 \geq 0 \Rightarrow |3x - 4| = 3x - 4 $
- $ x - 3 < 0 \Rightarrow |x - 3| = -x + 3 $

Inequality:
$$
3x - 4 \leq -x + 3 \\
4x \leq 7 \\
x \leq \frac{7}{4}
$$

Now intersect with $ \frac{4}{3} \leq x < 3 $:
- $ \frac{4}{3} \approx 1.333 $, $ \frac{7}{4} = 1.75 $

So:
$$
x \in \left[\frac{4}{3}, \frac{7}{4}\right]
$$

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#### Case 3: $ x \geq 3 $

Then:
- $ |3x - 4| = 3x - 4 $
- $ |x - 3| = x - 3 $

Inequality:
$$
3x - 4 \leq x - 3 \\
2x \leq 1 \\
x \leq \frac{1}{2}
$$

But $ x \geq 3 $ and $ x \leq \frac{1}{2} $ → no solution.

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#### Combine all valid parts:

From Case 1: $ \left[\frac{1}{2}, \frac{4}{3}\right) $
From Case 2: $ \left[\frac{4}{3}, \frac{7}{4}\right] $

Combined: $ \left[\frac{1}{2}, \frac{7}{4}\right] $

✔ Solution: $ \boxed{\left[\frac{1}{2}, \frac{7}{4}\right]} $

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2. $ 7 - 3|4x + 7| \leq -2 $

Solve algebraically.

$$
7 - 3|4x + 7| \leq -2 \\
-3|4x + 7| \leq -9 \\
|4x + 7| \geq 3
$$

Now solve $ |A| \geq B $ where $ B > 0 $:
$ A \leq -B $ or $ A \geq B $

So:
$$
4x + 7 \leq -3 \quad \text{or} \quad 4x + 7 \geq 3 \\
4x \leq -10 \quad \text{or} \quad 4x \geq -4 \\
x \leq -\frac{5}{2} \quad \text{or} \quad x \geq -1
$$

✔ Solution: $ \boxed{(-\infty, -\frac{5}{2}] \cup [-1, \infty)} $

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3. $ |x - 3| + |x + 2| < 13 $

Critical points: $ x = 3 $, $ x = -2 $

Divide into intervals:
- $ (-\infty, -2) $
- $ [-2, 3) $
- $ [3, \infty) $

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#### Case 1: $ x < -2 $

Then:
- $ x - 3 < 0 \Rightarrow |x - 3| = -(x - 3) = -x + 3 $
- $ x + 2 < 0 \Rightarrow |x + 2| = -(x + 2) = -x - 2 $

Sum:
$$
(-x + 3) + (-x - 2) = -2x + 1 < 13 \\
-2x < 12 \\
x > -6
$$

Intersect with $ x < -2 $: $ x \in (-6, -2) $

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#### Case 2: $ -2 \leq x < 3 $

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

Sum:
$$
(-x + 3) + (x + 2) = 5 < 13 \quad \text{Always true}
$$

So all $ x \in [-2, 3) $ satisfy.

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#### Case 3: $ x \geq 3 $

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

Sum:
$$
(x - 3) + (x + 2) = 2x - 1 < 13 \\
2x < 14 \\
x < 7
$$

Intersect with $ x \geq 3 $: $ x \in [3, 7) $

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Combine all:
- $ (-6, -2) $
- $ [-2, 3) $
- $ [3, 7) $

→ Union: $ (-6, 7) $

✔ Solution: $ \boxed{(-6, 7)} $

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4. $ |x + 3| > |2x - 1| $

Again, compare two absolute values.

Critical points: $ x = -3 $, $ x = \frac{1}{2} $

Intervals:
- $ (-\infty, -3) $
- $ [-3, \frac{1}{2}) $
- $ [\frac{1}{2}, \infty) $

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#### Case 1: $ x < -3 $

Then:
- $ x + 3 < 0 \Rightarrow |x + 3| = -x - 3 $
- $ 2x - 1 < 0 \Rightarrow |2x - 1| = -2x + 1 $

Inequality:
$$
-x - 3 > -2x + 1 \\
x > 4
$$

But $ x < -3 $, so no solution.

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#### Case 2: $ -3 \leq x < \frac{1}{2} $

Then:
- $ |x + 3| = x + 3 $
- $ |2x - 1| = -2x + 1 $

Inequality:
$$
x + 3 > -2x + 1 \\
3x > -2 \\
x > -\frac{2}{3}
$$

Intersect with $ [-3, \frac{1}{2}) $: $ x \in \left(-\frac{2}{3}, \frac{1}{2}\right) $

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#### Case 3: $ x \geq \frac{1}{2} $

Then:
- $ |x + 3| = x + 3 $
- $ |2x - 1| = 2x - 1 $

Inequality:
$$
x + 3 > 2x - 1 \\
3 + 1 > x \\
x < 4
$$

Intersect with $ x \geq \frac{1}{2} $: $ x \in \left[\frac{1}{2}, 4\right) $

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Now combine:
- $ \left(-\frac{2}{3}, \frac{1}{2}\right) $
- $ \left[\frac{1}{2}, 4\right) $

→ $ \left(-\frac{2}{3}, 4\right) $

✔ Solution: $ \boxed{\left(-\frac{2}{3}, 4\right)} $

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5. $ |3x - 3| + 5 > 14 $

Simplify:
$$
|3x - 3| > 9 \\
|3(x - 1)| > 9 \\
3|x - 1| > 9 \\
|x - 1| > 3
$$

So:
$$
x - 1 > 3 \quad \text{or} \quad x - 1 < -3 \\
x > 4 \quad \text{or} \quad x < -2
$$

✔ Solution: $ \boxed{(-\infty, -2) \cup (4, \infty)} $

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6. $ |x - 1| \geq 6 $

Standard form:
$$
x - 1 \geq 6 \quad \text{or} \quad x - 1 \leq -6 \\
x \geq 7 \quad \text{or} \quad x \leq -5
$$

✔ Solution: $ \boxed{(-\infty, -5] \cup [7, \infty)} $

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7. $ |2x + 2| \leq 12 $

Factor:
$$
|2(x + 1)| \leq 12 \\
2|x + 1| \leq 12 \\
|x + 1| \leq 6
$$

So:
$$
-6 \leq x + 1 \leq 6 \\
-7 \leq x \leq 5
$$

✔ Solution: $ \boxed{[-7, 5]} $

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8. $ |x - 2| < 9 $

This means:
$$
-9 < x - 2 < 9 \\
-7 < x < 11
$$

✔ Solution: $ \boxed{(-7, 11)} $

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## ✔ Final Answers (Summary):

1. $ \left[\frac{1}{2}, \frac{7}{4}\right] $
2. $ (-\infty, -\frac{5}{2}] \cup [-1, \infty) $
3. $ (-6, 7) $
4. $ \left(-\frac{2}{3}, 4\right) $
5. $ (-\infty, -2) \cup (4, \infty) $
6. $ (-\infty, -5] \cup [7, \infty) $
7. $ [-7, 5] $
8. $ (-7, 11) $

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Let me know if you'd like graphs for any of these!