Solve the inequality |x + 1| + |x + 2| > 3.

Step-by-step solution

1. 1

   Find the kink points. Each absolute value changes formula where its inside is zero:

   $x+1=0\Rightarrow x=-1,\qquad x+2=0\Rightarrow x=-2$

   These two points cut the line into three intervals: $x<-2$, $-2\le x<-1$, and $x\ge-1$. On each one, both absolute values can be replaced by plain linear expressions.

2. 2

   Case 1: x < -2 (both insides negative). Then $|x+1|=-x-1$ and $|x+2|=-x-2$, so

   $-2x-3>3\Rightarrow -2x>6\Rightarrow x<-3$

   Dividing by $-2$ flips the inequality. Intersecting $x<-3$ with the case condition $x<-2$ leaves $x<-3$.

3. 3

   Case 2: -2 <= x < -1 (mixed signs). Here $|x+1|=-x-1$ but $|x+2|=x+2$, and the $x$ terms cancel:

   $(-x-1)+(x+2)=1>3$

   This is false for every $x$ in the interval, so the middle strip contributes nothing. Geometrically, the sum of the two distances is constant at $1$ between the kinks.

4. 4

   Case 3: x >= -1 (both insides non-negative). Then

   $(x+1)+(x+2)=2x+3>3\Rightarrow 2x>0\Rightarrow x>0$

   Intersecting with $x\ge-1$ leaves $x>0$.

5. 5

   Union the cases.

   $x<-3\quad\text{or}\quad x>0,\qquad\text{i.e.}\qquad(-\infty,-3)\cup(0,\infty)$

6. 6

   Read the answer as distances, then verify. $|x+1|+|x+2|$ is the total distance from $x$ to $-1$ and to $-2$; its minimum is the gap $1$, reached on the whole middle interval, and it grows by $2$ per unit outside — so it hits $3$ exactly one unit outside each kink, at $x=-3$ and $x=0$. Checking: $|{-2}|+|{-1}|=3$ at $x=-3$ ✓ and $|1|+|2|=3$ at $x=0$ ✓, and a scan of $1601$ test points from $-8$ to $8$ agrees with the stated solution set at every point ✓.