SOLUTION: Solving absolute value equations and inequalities ... - Free Printable

You’ve uploaded a worksheet titled “Absolute Value Equations and Inequalities” that explains how to solve linear absolute value equations, with definitions, steps, and examples. Since you asked to “solve the problem and explain the solution,” I’ll assume you want me to explain the entire concept shown in the image, including the logic behind each example — as if you’re learning this for the first time.

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## 📘 Absolute Value Definition

The absolute value of a number \( x \), written \( |x| \), is its distance from zero on the number line. It’s always non-negative.

\[
|x| =
\begin{cases}
x & \text{if } x \geq 0 \\
-x & \text{if } x < 0
\end{cases}
\]

> 💡 Think of it like this:
> - If \( x \) is positive or zero, its absolute value is itself.
> - If \( x \) is negative, its absolute value is the opposite (positive version).

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## ✔ Steps for Solving Linear Absolute Value Equations: \( |ax + b| = c \)

Step 1: Isolate the absolute value expression.

Move all other terms to the other side so you have something like:
\[
|expression| = number
\]

Step 2: Analyze the right-hand side (RHS):

#### Case A: RHS = 0 → One Solution
Since only 0 has an absolute value of 0,
\[
|expression| = 0 \Rightarrow expression = 0
\]
Solve that single equation.

#### Case B: RHS < 0 → No Solution
Absolute value can never be negative. So if you get:
\[
|expression| = -5 \quad \text{(or any negative number)}
\]
→ No solution exists.

#### Case C: RHS > 0 → Two Solutions
If the RHS is positive, then the expression inside the absolute value could be either:
- Equal to the positive number, OR
- Equal to the negative of that number.

So you write:
\[
expression = c \quad \text{OR} \quad expression = -c
\]
Then solve both equations separately.

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## 🔍 Let’s Walk Through Each Example:

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Example a: \( |3x + 12| + 7 = 7 \)

✔ Step 1: Isolate the absolute value.

Subtract 7 from both sides:
\[
|3x + 12| = 0
\]

✔ Step 2: RHS = 0 → One solution.

Set inside equal to zero:
\[
3x + 12 = 0
\]
\[
3x = -12
\]
\[
x = -4
\]

✔ Final Answer: \( \boxed{x = -4} \)

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Example b: \( |3x - 7| + 7 = 2 \)

✔ Step 1: Isolate the absolute value.

Subtract 7:
\[
|3x - 7| = -5
\]

✔ Step 2: RHS = -5 → Negative → No solution

Absolute value cannot be negative. So no real number satisfies this.

✔ Final Answer: \( \boxed{\text{No Solution}} \)

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Example c: \( |3x - 7| + 7 = 9 \)

✔ Step 1: Isolate the absolute value.

Subtract 7:
\[
|3x - 7| = 2
\]

✔ Step 2: RHS = 2 → Positive → Two solutions.

Set up two equations:

1. \( 3x - 7 = 2 \)
\[
3x = 9 \Rightarrow x = 3
\]

2. \( 3x - 7 = -2 \)
\[
3x = 5 \Rightarrow x = \frac{5}{3}
\]

✔ Final Answer: \( \boxed{x = 3} \quad \text{or} \quad \boxed{x = \frac{5}{3}} \)

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Example d: \( |x + 5| = |2x - 1| \)

This is different — both sides are absolute values.

✔ Strategy: When two absolute values are equal, their insides are either equal OR opposites.

So set up two cases:

#### Case 1: Insides are equal
\[
x + 5 = 2x - 1
\]
\[
5 + 1 = 2x - x
\]
\[
6 = x
\]

#### Case 2: Insides are opposites
\[
x + 5 = -(2x - 1)
\]
Distribute the negative:
\[
x + 5 = -2x + 1
\]
Add \( 2x \) to both sides:
\[
3x + 5 = 1
\]
Subtract 5:
\[
3x = -4
\]
\[
x = -\frac{4}{3}
\]

✔ Final Answer: \( \boxed{x = 6} \quad \text{or} \quad \boxed{x = -\frac{4}{3}} \)

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## 🧠 Key Takeaways:

- Absolute value = distance → always ≥ 0.
- If \( |A| = 0 \) → \( A = 0 \)
- If \( |A| = \text{negative} \) → impossible → no solution
- If \( |A| = \text{positive} \) → two possibilities: \( A = \text{pos} \) or \( A = \text{neg} \)
- If \( |A| = |B| \) → \( A = B \) or \( A = -B \)

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## ✔ Practice Tip:

Always check your answers by plugging them back into the original equation!

For example, in Example d:
- Plug \( x = 6 \): \( |6+5| = |12-1| → |11| = |11| → 11=11 ✔️ \)
- Plug \( x = -\frac{4}{3} \): \( |-\frac{4}{3}+5| = |-\frac{8}{3}-1| → |\frac{11}{3}| = |-\frac{11}{3}| → \frac{11}{3} = \frac{11}{3} ✔️ \)

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Let me know if you’d like to try more problems or need help with inequalities next! 😊