To solve the inequality [tex]\( |x + 1| > 2 \)[/tex], we need to consider the properties of absolute values.
The absolute value [tex]\( |x+1| \)[/tex] measures the distance of [tex]\( x+1 \)[/tex] from 0 on the number line, and is defined such that:
[tex]\[
|x+1| =
\begin{cases}
x + 1 & \text{if } x + 1 \ge 0 \\
-(x + 1) & \text{if } x + 1 < 0
\end{cases}
\][/tex]
Given the inequality [tex]\( |x + 1| > 2 \)[/tex], we can break it down into two separate inequalities:
[tex]\[
x + 1 > 2 \quad \text{or} \quad x + 1 < -2
\][/tex]
Let's solve each inequality separately:
1. For the inequality [tex]\( x + 1 > 2 \)[/tex]:
[tex]\[
x + 1 > 2 \implies x > 2 - 1 \implies x > 1
\][/tex]
2. For the inequality [tex]\( x + 1 < -2 \)[/tex]:
[tex]\[
x + 1 < -2 \implies x < -2 - 1 \implies x < -3
\][/tex]
By combining these two results, we obtain the solution set:
[tex]\[
x < -3 \quad \text{or} \quad x > 1
\][/tex]
This means that [tex]\( x \)[/tex] can be any value less than -3 or any value greater than 1.
Graphically, this solution is represented as two rays on the number line, one extending to the left from -3 (not including -3 itself), and the other extending to the right from 1 (not including 1 itself).
Let's look at the provided options:
A. Solution: [tex]\( x < -3 \)[/tex] or [tex]\( x > 1 \)[/tex]
B. Solution: [tex]\( x > -3 \)[/tex] and [tex]\( x < 1 \)[/tex]
C. Solution: [tex]\( x < -3 \)[/tex] or [tex]\( x > 1 \)[/tex]
D. Solution: [tex]\( x > -1 \)[/tex] and [tex]\( x < 3 \)[/tex]
Both options A and C provide the correct solution [tex]\( x < -3 \)[/tex] or [tex]\( x > 1 \)[/tex].
Therefore, the correct answer is either:
A. Solution: [tex]\( x < -3 \)[/tex] or [tex]\( x > 1 \)[/tex]
or
C. Solution: [tex]\( x < -3 \)[/tex] or [tex]\( x > 1 \)[/tex]
Since both A and C represent the correct solution and graph correctly, they are equivalent for the purposes of this problem.
