Solve the equation $|2x-1| -|x+5| = 3$ 
Problem : Solve the equation $|2x-1| - |x+5| = 3$

In my attempt to solve the problem, I only manage to get one of the solutions.
Attempted Solution
$$\begin{equation}
\begin{split}
|x|-|y| & \leq |x-y| \\
\implies |2x-1| - |x+5| & \leq |(2x-1)-(x+5)| \\
\implies 3 &\leq |x-6| \\
\implies 3 &= |x-6| \\
\implies (x = 9) \ & \lor \ (x= 3 \ (\text{Reject})) \\
\implies x &= 9
\end{split}
\end{equation}
$$
However plugging the correct solution set to this equation is $x \in \{-\frac{7}{3}, 9\}$. Why does my attempted solution, not reach $-\frac{7}{3}$, as one of the possible solutions? Furthermore, how would you go about solving a problem of this nature?
 A: Remember that $|x|=\begin{cases}x&\text{if}~x\geq 0\\ -x&\text{if}~x<0\end{cases}$
Now, we attempt to rewrite the original expression without absolute value signs.  To do so, we ask ourselves when do we need to change the sign for each?  On what regions will we need to change the sign of $|2x-1|$?  The sign of $|x+5|$?
$2x-1<0$ when $x<\frac{1}{2}$.
$x+5<0$ when $x<-5$
We see then that the expression can be written without absolute value signs if broken into the following three regions: case 1:$x<-5$, case 2: $-5\leq x<\frac{1}{2}$, and case 3: $\frac{1}{2}\leq x$
$|2x-1|-|x+5| = \begin{cases} -(2x-1)-(-(x+5))&\text{when}~x<-5\\
-(2x-1)-(x+5)&\text{when}~-5\leq x<\frac{1}{2}\\
(2x-1)-(x+5)&\text{when}~\frac{1}{2}\leq x\end{cases}$
Setting each of these equal to three and solving for $x$ individually, and then checking that such a value for $x$ lies within its respective region will finish the problem.
A: Because $|2x-1|-|x+5|$ is also $\leq |(x+5) - (2x-1)|$.  (Multiplying through by $-1$ doesn't change the absolute value...)
The generic method is this:


*

*$x+5$ is nonnegative for $x \geq -5$ and negative for $x < -5$.

*$2x-1$ is nonnegative for $x \geq 1/2$ and negative for $x < 1/2$.

*On the interval $(-\infty,-5)$, both are negative, so we solve  $$ |2x-1|-|x+5| = -(2x-1)--(x+5) = -x + 6 = 3  \text{,}  $$ so $x = 3$, but this is not in $(-\infty, -5)$, so we reject it.

*On the interval $[-5,1/2)$, we solve $$ |2x-1|-|x+5| = -(2x-1)-(x+5) = -3x - 4 = 3  \text{,}  $$ so $x = -7/3$, which is in $[-5,1/2)$.

*On the interval $[1/2,\infty)$, we solve $$ |2x-1|-|x+5| = (2x-1)-(x+5) = x - 6 = 3  \text{,}  $$ so $x = 9$, which is in $[1/2,\infty)$.


So the solution set is $\{-7/3,9\}$.
A: The expressions $2x-1$ and $x+5$ can take on positive, zero, and negative values.
You considered the case for which both expressions were positive.
If $-5 < x < 1/2$, then one is positive and one is negative, so your equation for those values of $x$ becomes:
$$(1-2x) - (x+5) = 3$$
If $x < -5$ then both expressions are negative and you have
$$(1-2x) - (-x -5) = 3$$
This equation doesn't produce a valid solution, because even though $x=3$ solves the equation, it's not less than $-5$.
The equation above, though, has $x = -7/3$ as a solution, and as it's between $1/2$ and $-5$, it also solve the original equation.
