If `|x+3|ge10,` then:

Options

A. `x in(-13,7)`
B. `(-13,7)`
C. `x in(-oo,-13]cup[7,oo)`
D. `x in(-oo,-13)cup(7,oo)`

Answer

To solve the inequality \( |x + 3| \geq 10 \), we will break it down into two cases based on the definition of absolute value.

### Step 1: Set up the cases
The expression \( |a| \geq b \) implies two cases:
1. \( a \geq b \)
2. \( a \leq -b \)

In our case, \( a = x + 3 \) and \( b = 10 \). Therefore, we can write:
1. \( x + 3 \geq 10 \)
2. \( x + 3 \leq -10 \)

### Step 2: Solve the first case
For the first case:
\[ x + 3 \geq 10 \]
Subtract 3 from both sides:
\[ x \geq 10 - 3 \]
\[ x \geq 7 \]

### Step 3: Solve the second case
For the second case:
\[ x + 3 \leq -10 \]
Subtract 3 from both sides:
\[ x \leq -10 - 3 \]
\[ x \leq -13 \]

### Step 4: Combine the results
From the two cases, we have:
1. \( x \geq 7 \)
2. \( x \leq -13 \)

This means the solution can be expressed in interval notation as:
\[ x \in (-\infty, -13] \cup [7, \infty) \]

### Final Answer
The solution to the inequality \( |x + 3| \geq 10 \) is:
\[ x \in (-\infty, -13] \cup [7, \infty) \]

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