# Find all real numbers $x$ such that $||x+1| - |3x - 1|| < 1$

This is on my final exam review. I have the solution, but I do not understand it. When looking at $|x+1| - |3x - 1|$, I see four cases:

a) $x + 1 > 0$ and $3x - 1 > 0$

$x > -1$ and $x > \frac{1}{3}$

$x > -1$

b) $x + 1 < 0$ and $3x - 1 > 0$

$x < -1$ and $x > \frac{1}{3}$ (cannot happen)

c) $x + 1 > 0$ and $3x - 1 < 0$

$x > -1$ and $x < \frac{1}{3}$

$-1 < x < \frac{1}{3}$

d) $x + 1 < 0$ and $3x - 1 < 0$

$x < -1$ and $x < \frac{1}{3}$

$x < \frac{1}{3}$

So when solving the equation, I would look at cases a, c, and d. However, the solution says to use $x < -1$, $-1 \leq x < \frac{1}{3}$, and $\frac{1}{3} \leq x$

What am I missing here?

-

Case (a) is $x\gt\frac13$, case (b) cannot happen, case (c) is $-1\lt x\lt\frac13$ and case (d) is $x\lt-1$.
In case (d), [$x\lt-1$ and $x\lt\frac13$] is equivalent to [$x\lt-1$], not to [$x\lt\frac13$]. In case (a), [$x\gt-1$ and $x\gt\frac13$] is equivalent to [$x\gt\frac13$], not to [$x\gt-1$].