Solve the following systems of inequalities for all `x in R` Solve for real x, |x+1|+|x|>3

Answer

To solve the inequality \( |x+1| + |x| > 3 \), we will break it down into different cases based on the values of \( x \).

### Step 1: Identify critical points
The critical points for the absolute values are found by setting the expressions inside the absolute values to zero:
- \( x + 1 = 0 \) gives \( x = -1 \)
- \( x = 0 \)

These points divide the number line into three intervals:
1. \( (-\infty, -1) \)
2. \( [-1, 0) \)
3. \( [0, \infty) \)

### Step 2: Case 1: \( x < -1 \)
In this interval, both \( x + 1 \) and \( x \) are negative, so:
\[ |x + 1| = -(x + 1) = -x - 1 \]
\[ |x| = -x \]
Substituting these into the inequality:
\[ -x - 1 - x > 3 \]
This simplifies to:
\[ -2x - 1 > 3 \]
Adding 1 to both sides:
\[ -2x > 4 \]
Dividing by -2 (remember to reverse the inequality):
\[ x < -2 \]
Thus, for this case, the solution is \( x < -2 \).

### Step 3: Case 2: \( -1 \leq x < 0 \)
In this interval, \( x + 1 \) is non-negative and \( x \) is negative:
\[ |x + 1| = x + 1 \]
\[ |x| = -x \]
Substituting these into the inequality:
\[ x + 1 - x > 3 \]
This simplifies to:
\[ 1 > 3 \]
This is not valid, so there are no solutions in this interval.

### Step 4: Case 3: \( x \geq 0 \)
In this interval, both \( x + 1 \) and \( x \) are non-negative:
\[ |x + 1| = x + 1 \]
\[ |x| = x \]
Substituting these into the inequality:
\[ x + 1 + x > 3 \]
This simplifies to:
\[ 2x + 1 > 3 \]
Subtracting 1 from both sides:
\[ 2x > 2 \]
Dividing by 2:
\[ x > 1 \]
Thus, for this case, the solution is \( x > 1 \).

### Step 5: Combine the solutions
From the three cases, we have:
- From Case 1: \( x < -2 \)
- From Case 2: No solutions
- From Case 3: \( x > 1 \)

The overall solution set is:
\[ x \in (-\infty, -2) \cup (1, \infty) \]

### Final Answer
The solution to the inequality \( |x + 1| + |x| > 3 \) is:
\[ x \in (-\infty, -2) \cup (1, \infty) \]