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,
 A: This is how I learned it back in Algebra I. 
You can split it into two equations. Since the sign is $\le$, it will be that Equation 1 and Equation 2 are true.
Here are the steps: 
$$|3-5x| \le x$$

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

$$6x \ge 3  \;\;\;\;\;\;\;| \;\;-4x \ge -3$$

$$x \ge \frac 12  \;\;\;| \;\;\;\;x \le \frac 34$$

$$\frac 12 \le x \le \frac 34$$
A: Divide $ -x \leq 3-5x \leq x $ by $x$ and go from there. Although I think what you have is correct, the solution IS the interval $ \dfrac{1}{2} \leq x \leq \dfrac{3}{4}$.
A: 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: $\frac{1}{2} \leq x \leq \frac{3}{4}$.
A: Your solution was correct except you needed to use the word AND correctly.
The inequality $-x \le 3-5x \le x$ is equivalent to the inequalities
\begin{align}
   -x \le 3-5x \quad &\text{AND} \quad 3-5x \le x \\
   4x \le 3 \quad &\text{AND} \quad 3 \le 6x \\
   x \le \frac 34 \quad &\text{AND} \quad \frac 12 \le x \\
   \frac 12 \le x \quad &\text{AND} \quad x \le \frac 34 \\
x &\in\left[\frac 12, \frac 34 \right]
\end{align}
