Solving Absolute Value Equations and Inequalities WS | PDF ... - Free Printable

To solve the given absolute value equations, we need to consider the definition of absolute value. The absolute value equation \( |A| = B \) implies two cases:

1. \( A = B \)
2. \( A = -B \)

We will apply this principle to each problem step by step.

---

Problem a: \( |x - 6| = 8 \)

#### Step 1: Set up the two cases
\[ x - 6 = 8 \quad \text{or} \quad x - 6 = -8 \]

#### Step 2: Solve each case
1. \( x - 6 = 8 \)
\[
x = 8 + 6 \implies x = 14
\]

2. \( x - 6 = -8 \)
\[
x = -8 + 6 \implies x = -2
\]

#### Step 3: Check the solutions
- For \( x = 14 \):
\[
|14 - 6| = |8| = 8 \quad \text{(True)}
\]
- For \( x = -2 \):
\[
|-2 - 6| = |-8| = 8 \quad \text{(True)}
\]

#### Final Answer:
\[
\boxed{x = 14 \text{ or } x = -2}
\]

---

Problem b: \( |x + 2| = 8 \)

#### Step 1: Set up the two cases
\[ x + 2 = 8 \quad \text{or} \quad x + 2 = -8 \]

#### Step 2: Solve each case
1. \( x + 2 = 8 \)
\[
x = 8 - 2 \implies x = 6
\]

2. \( x + 2 = -8 \)
\[
x = -8 - 2 \implies x = -10
\]

#### Step 3: Check the solutions
- For \( x = 6 \):
\[
|6 + 2| = |8| = 8 \quad \text{(True)}
\]
- For \( x = -10 \):
\[
|-10 + 2| = |-8| = 8 \quad \text{(True)}
\]

#### Final Answer:
\[
\boxed{x = 6 \text{ or } x = -10}
\]

---

Problem c: \( |-9x| = 64 \)

#### Step 1: Simplify the absolute value
Since \( |-9x| = |9x| \), we have:
\[ |9x| = 64 \]

#### Step 2: Set up the two cases
\[ 9x = 64 \quad \text{or} \quad 9x = -64 \]

#### Step 3: Solve each case
1. \( 9x = 64 \)
\[
x = \frac{64}{9}
\]

2. \( 9x = -64 \)
\[
x = -\frac{64}{9}
\]

#### Step 4: Check the solutions
- For \( x = \frac{64}{9} \):
\[
|-9 \cdot \frac{64}{9}| = |-64| = 64 \quad \text{(True)}
\]
- For \( x = -\frac{64}{9} \):
\[
|-9 \cdot -\frac{64}{9}| = |64| = 64 \quad \text{(True)}
\]

#### Final Answer:
\[
\boxed{x = \frac{64}{9} \text{ or } x = -\frac{64}{9}}
\]

---

Problem d: \( |-7x + 4| = 18 \)

#### Step 1: Set up the two cases
\[ -7x + 4 = 18 \quad \text{or} \quad -7x + 4 = -18 \]

#### Step 2: Solve each case
1. \( -7x + 4 = 18 \)
\[
-7x = 18 - 4 \implies -7x = 14 \implies x = -2
\]

2. \( -7x + 4 = -18 \)
\[
-7x = -18 - 4 \implies -7x = -22 \implies x = \frac{22}{7}
\]

#### Step 3: Check the solutions
- For \( x = -2 \):
\[
|-7(-2) + 4| = |14 + 4| = |18| = 18 \quad \text{(True)}
\]
- For \( x = \frac{22}{7} \):
\[
|-7 \cdot \frac{22}{7} + 4| = |-22 + 4| = |-18| = 18 \quad \text{(True)}
\]

#### Final Answer:
\[
\boxed{x = -2 \text{ or } x = \frac{22}{7}}
\]

---

Problem e: \( |4x + 4| = 28 \)

#### Step 1: Set up the two cases
\[ 4x + 4 = 28 \quad \text{or} \quad 4x + 4 = -28 \]

#### Step 2: Solve each case
1. \( 4x + 4 = 28 \)
\[
4x = 28 - 4 \implies 4x = 24 \implies x = 6
\]

2. \( 4x + 4 = -28 \)
\[
4x = -28 - 4 \implies 4x = -32 \implies x = -8
\]

#### Step 3: Check the solutions
- For \( x = 6 \):
\[
|4(6) + 4| = |24 + 4| = |28| = 28 \quad \text{(True)}
\]
- For \( x = -8 \):
\[
|4(-8) + 4| = |-32 + 4| = |-28| = 28 \quad \text{(True)}
\]

#### Final Answer:
\[
\boxed{x = 6 \text{ or } x = -8}
\]

---

Problem f: \( 5|n + 10| = 10 \)

#### Step 1: Isolate the absolute value
\[
|n + 10| = \frac{10}{5} \implies |n + 10| = 2
\]

#### Step 2: Set up the two cases
\[ n + 10 = 2 \quad \text{or} \quad n + 10 = -2 \]

#### Step 3: Solve each case
1. \( n + 10 = 2 \)
\[
n = 2 - 10 \implies n = -8
\]

2. \( n + 10 = -2 \)
\[
n = -2 - 10 \implies n = -12
\]

#### Step 4: Check the solutions
- For \( n = -8 \):
\[
5|-8 + 10| = 5|2| = 5 \cdot 2 = 10 \quad \text{(True)}
\]
- For \( n = -12 \):
\[
5|-12 + 10| = 5|-2| = 5 \cdot 2 = 10 \quad \text{(True)}
\]

#### Final Answer:
\[
\boxed{n = -8 \text{ or } n = -12}
\]

---

Problem g: \( |l - 6r| + 3 = 46 \)

#### Step 1: Isolate the absolute value
\[
|l - 6r| + 3 = 46 \implies |l - 6r| = 46 - 3 \implies |l - 6r| = 43
\]

#### Step 2: Set up the two cases
\[ l - 6r = 43 \quad \text{or} \quad l - 6r = -43 \]

#### Step 3: Solve each case
1. \( l - 6r = 43 \)
\[
l = 43 + 6r
\]

2. \( l - 6r = -43 \)
\[
l = -43 + 6r
\]

#### Final Answer:
\[
\boxed{l = 43 + 6r \text{ or } l = -43 + 6r}
\]

---

Problem h: \( \frac{|3v - 2|}{5} = 4 \)

#### Step 1: Isolate the absolute value
\[
\frac{|3v - 2|}{5} = 4 \implies |3v - 2| = 4 \cdot 5 \implies |3v - 2| = 20
\]

#### Step 2: Set up the two cases
\[ 3v - 2 = 20 \quad \text{or} \quad 3v - 2 = -20 \]

#### Step 3: Solve each case
1. \( 3v - 2 = 20 \)
\[
3v = 20 + 2 \implies 3v = 22 \implies v = \frac{22}{3}
\]

2. \( 3v - 2 = -20 \)
\[
3v = -20 + 2 \implies 3v = -18 \implies v = -6
\]

#### Step 4: Check the solutions
- For \( v = \frac{22}{3} \):
\[
\frac{|3 \cdot \frac{22}{3} - 2|}{5} = \frac{|22 - 2|}{5} = \frac{|20|}{5} = \frac{20}{5} = 4 \quad \text{(True)}
\]
- For \( v = -6 \):
\[
\frac{|3(-6) - 2|}{5} = \frac{|-18 - 2|}{5} = \frac{|-20|}{5} = \frac{20}{5} = 4 \quad \text{(True)}
\]

#### Final Answer:
\[
\boxed{v = \frac{22}{3} \text{ or } v = -6}
\]

---

Final Answers for All Problems:

\[
\boxed{
\begin{aligned}
&\text{a. } x = 14 \text{ or } x = -2 \\
&\text{b. } x = 6 \text{ or } x = -10 \\
&\text{c. } x = \frac{64}{9} \text{ or } x = -\frac{64}{9} \\
&\text{d. } x = -2 \text{ or } x = \frac{22}{7} \\
&\text{e. } x = 6 \text{ or } x = -8 \\
&\text{f. } n = -8 \text{ or } n = -12 \\
&\text{g. } l = 43 + 6r \text{ or } l = -43 + 6r \\
&\text{h. } v = \frac{22}{3} \text{ or } v = -6
\end{aligned}
}
\]