Example of an absolute value equation |x + 3| = |x - 11| from a math tutorial.

We are given the absolute value equation:

$$
|x + 3| = |x - 11|
$$

Our goal is to solve for $ x $.

---

🔍 Step 1: Understand what this means

The absolute value of a number is its distance from zero on the number line. So, this equation says:

> The distance from $ x $ to $ -3 $ is equal to the distance from $ x $ to $ 11 $.

This suggests that $ x $ is equidistant from $ -3 $ and $ 11 $. That means $ x $ must be the midpoint between $ -3 $ and $ 11 $.

Let’s compute that midpoint:

$$
\text{Midpoint} = \frac{-3 + 11}{2} = \frac{8}{2} = 4
$$

So, one possible solution is $ x = 4 $.

But let's solve it algebraically to confirm and make sure we don't miss any cases.

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✔ Step 2: Solve Algebraically

We use the property of absolute values:

> If $ |A| = |B| $, then either:
> - $ A = B $, or
> - $ A = -B $

Apply this to our equation:

$$
|x + 3| = |x - 11|
$$

So we have two cases:

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#### ✔ Case 1: $ x + 3 = x - 11 $

Subtract $ x $ from both sides:

$$
3 = -11
$$

✘ This is false. So no solution in this case.

---

#### ✔ Case 2: $ x + 3 = -(x - 11) $

Simplify the right-hand side:

$$
x + 3 = -x + 11
$$

Add $ x $ to both sides:

$$
2x + 3 = 11
$$

Subtract 3:

$$
2x = 8
$$

Divide by 2:

$$
x = 4
$$

✔ This is a valid solution.

---

🚨 But wait — could there be more solutions?

Let’s consider the critical points where the expressions inside the absolute values change sign:

- $ x + 3 = 0 $ → $ x = -3 $
- $ x - 11 = 0 $ → $ x = 11 $

These divide the number line into three intervals:

1. $ x < -3 $
2. $ -3 \leq x < 11 $
3. $ x \geq 11 $

We can solve the equation in each interval by removing absolute values based on the sign in that region.

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🔹 Interval 1: $ x < -3 $

In this region:
- $ x + 3 < 0 $ → $ |x + 3| = -(x + 3) $
- $ x - 11 < 0 $ → $ |x - 11| = -(x - 11) $

So the equation becomes:

$$
-(x + 3) = -(x - 11)
$$

Simplify:

$$
-x - 3 = -x + 11
$$

Add $ x $ to both sides:

$$
-3 = 11
$$

✘ False. No solution in this interval.

---

🔹 Interval 2: $ -3 \leq x < 11 $

Here:
- $ x + 3 \geq 0 $ → $ |x + 3| = x + 3 $
- $ x - 11 < 0 $ → $ |x - 11| = -(x - 11) $

So:

$$
x + 3 = - (x - 11) = -x + 11
$$

Same as before:

$$
x + 3 = -x + 11 \\
2x = 8 \\
x = 4
$$

Now check if $ x = 4 $ is in this interval:
Yes, $ -3 \leq 4 < 11 $ → ✔ Valid.

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🔹 Interval 3: $ x \geq 11 $

Here:
- $ x + 3 > 0 $ → $ |x + 3| = x + 3 $
- $ x - 11 \geq 0 $ → $ |x - 11| = x - 11 $

So:

$$
x + 3 = x - 11
$$

Subtract $ x $:

$$
3 = -11
$$

✘ False. No solution here.

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✔ Final Answer:

The only solution is:

$$
\boxed{x = 4}
$$

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📌 Interpretation:

$ x = 4 $ is equidistant from $ -3 $ and $ 11 $:

- Distance to $ -3 $: $ |4 - (-3)| = |7| = 7 $
- Distance to $ 11 $: $ |4 - 11| = |-7| = 7 $

✔️ Confirmed!

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✔ Summary:

$$
\boxed{x = 4}
$$

is the only solution to the equation $ |x + 3| = |x - 11| $.