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,

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: $\frac{1}{2} \leq x \leq \frac{3}{4}$.
