Find the solutions to the inequality: $$|x+1| \geq 3$$
I translate this as: which numbers are at least $3$ units from $1$? So, picturing a number line, I would place a filled in circle at the point $1$. The solutions would then be on the interval $(-\infty,-2] \cup [4,\infty)$. But this is wrong, because:
Why do they rewrite $|x+1|$ as $|x-(-1)|$?
On
You could think about it like this:
If you're solving for $|x+1|\ge3$, take away the absolute value sign and you'll have $x+1\ge3$. Solve for $x$: you'll get $x\ge2$, which we can see is part of the answer. Now we're missing the interval $(-\infty,-4]$, how do we get that? Well, the thing that I didn't mention earlier, which should jump out to you, is that $|-x-1|=|x+1|$ (think about it, if you have $x=5$, for example, $|5+1|=6, |-5-1|=6$ as well), so $|x+1|$ could be $x+1$ or $-x-1$, so you have to solve both inequalities (namely you have to solve both $x+1\ge3$ and $-x-1\ge3$)!
$-x-1\ge3, -x\ge4$, multiply by $-1$ so we have a positive $x$ on the left side. We then get $x\le-4$ (remember, we have to switch the sign of the inequality when you multiply/divide by a negative number. $x\le-4$ is the same as $(-\infty, -4]$. Hope this helps!
The difference $|a-b|$ signifies the positive distance between two numbers $a,b$ on the real line. So when you consider $|x+1|$, this doesn't mean the distance between $x$ and $1$, because it isn't a difference, but instead the distance between $x$ and $-1$, since $|x+1|=|x-(-1)|$.