<s>[INST] 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?
 [/INST] 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 $\{-7/3,9\}$.
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