<s>[INST] Absolute value inequality with variable on both sides I am trying to solve the following inequality:

$$|3-5x| \le x$$

I am not familiar with inequalities including one absolute value with variables on both sides. I tried to solve it as follows:

$$-x \le 3-5x \le x$$

Then I solved for each side separately,as follows:

$$3-5x \le x$$
$$  x \ge (1/2)$$



$$-x \le 3-5x$$
$$x \le \frac34$$

I know my solution is incorrect and that it actually lies between $\frac12$ and $\frac34$, but I wanted to know what is wrong with my method and what is the appropriate approach to solving such inequalities.
Thanks,
 [/INST] You need to break up the absolute value into its intervals:
$$
|x| = \begin{cases}
x & x > 0 \\
-x & x < 0 \\
0 & x = 0
\end{cases}
$$
Therefore for $|3-5x|$ you need to find the interval when it's less than zero and when it's greater than zero:
$$
|3-5x| = \begin{cases}
3 - 5x & 3 - 5x > 0 \rightarrow 3 > 5x \rightarrow x < \frac{3}{5}\\
5x - 3 & 3 - 5x < 0 \rightarrow 3 <5x \rightarrow x > \frac{3}{5} \\
0 & 3 - 5x = 0 \rightarrow 3 = 5x \rightarrow x = \frac{3}{5}
\end{cases}
$$
Now you solve the inequality in each case:


*

*$x < \frac{3}{5} \rightarrow |3 - 5x| = 3 - 5x$
$$
3 - 5x \leq x \\
3 \leq 6x \\
x \geq \frac{1}{2}
$$

*$x > \frac{3}{5} \rightarrow |3 - 5x| = 5x - 3$
$$
5x - 3 \leq x \\
4x \leq 3 \\
x \leq \frac{3}{4}
$$

*$x = \frac{3}{5} \rightarrow |3 - 5x| = 0$
$$
0 \leq x \\
x \geq 0
$$


Now you need to analyze each case:


*

*$x < \frac{3}{5} \wedge x \geq \frac{1}{2}$


It is true that $\frac{3}{5} = \frac{6}{10} \geq \frac{5}{10}$.  Therefore this particular interval is true for $\frac{1}{2} \leq x < \frac{3}{5}$.


*$x > \frac{3}{5} \wedge x \leq \frac{3}{4}$


Since $\frac{3}{5} = \frac{12}{20}$ and $\frac{3}{4}  = \frac{15}{20}$, this is true for $\frac{3}{5} < x \leq \frac{3}{4}$.


*$x = \frac{3}{5} \wedge x \geq 0$


$\frac{3}{5} > 0$ therefore this is trivially satisfied--thus $x = \frac{3}{5}$ is allowed.
When we combine these results, we find that $\frac{1}{2} \leq x \leq \frac{3}{4}$.
PS Edit:
It's probably easier to use:
$$
|x| = \begin{cases}
x & x \geq 0 \\
-x & x \leq 0 \\
\end{cases}
$$
In that case you get the two intervals:
$$
\frac{1}{2} \leq x \leq \frac{3}{5}
$$
and 
$$
\frac{3}{5} \leq x \leq \frac{3}{4}
$$
Which clearly combines to give: $\frac{1}{2} \leq x \leq \frac{3}{4}$.
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