If $x$ is a real number, then $|x+1| \leq 3$ implies that $-4 \leq x \leq 2$. 
If $x$ is a real number, then $|x+1| \leq 3$ implies that $-4 \leq x \leq 2$.

I've tried to prove this by exhaustion, is that the right way to prove it? 
 A: If $a$ and $b$ are real numbers and if $b \ge 0$, then the inequality $|a| \le b$ exactly means
$$-b \le a \le b.$$
Hence , if $b=3$, then
$$|a| \le 3 \iff -3 \le a \le 3.$$
Now let $a=x+1$. Then we get
$$|x+1| \le 3 \iff -3 \le x+1 \le 3 \iff -4 \le x \le 2.$$
A: You can use the definition of $|.|$.
The absolute value of real number $a$ (we'll write it $|a|$) it's the distance between the point, 
which corresponds to $a$ in the $x$-axis and the origin.
In our case the distance between $x+1$ and $0$ less or equal to $3$.
Thus, we have $$-3\leq x+1\leq3$$ or
$$-4\leq x\leq2$$
A: For real valued $x$ we have two cases.


*

*Either $x+1\geq 0$, in which case $\lvert x+1\rvert =(x+1)$ so...  $0\leq x+1 \leq 3$

*Or $x+1<0$, in which case $\lvert x+1\rvert=-(x+1)$, so... $0 < -(x+1)\leq 3$.
Thus $(\lvert x+1\rvert \leq 3 )\iff ( -3\leq x+1 \leq 3 )$
A: A good thing to think about expressions like $| x+ a| \leq b $ is that they are equivalent to saying $|x - (-a)| \leq b$, or that the 'distance' between $x$ and $-a$ is less than $b$. Then in the context of your problem it is clear that $x$ lies between the bounds given.
A: I'd say that the easiest way to prove this is to simply solve the inequality for x.
We know that:
|x+1| = 3, simplifies to:
x+1 = +3 AND x+1=-3       
So by this logic, we can say that:
x+1 ≤ 3  AND x+1 ≥ -3  (switching the ≤ to ≥ when we change the sign on the 3)
x ≤ 2    AND x ≥ -4    (subtracting 1 in on both sides)
Then we can just simplify this expression down into the one, written as, -4 ≤ x ≤ 2, just as you needed to show.
Hope this helps. This is just the easiest way to show it, though there are many others.
