Any tips for solving $\frac{|4x-2|}{|2x+1|} \le 1$ as succinctly as possible? $\frac{|4x-2|}{|2x+1|} \le 1$
So as I currently see it, I have two choices:
1) Attempt to solve algebraically but that has led me down some long paths when I believe the question should be solvable rather more quickly
2) Draw graphs... But owing to the unpredictable complexity of the equation (I'm sitting an exam soon), I'd rather not bank on that option.
Any help would be great!
 A: The quickest and simplest way is to look where $-2x-1\leq 4x-2\leq 2x+1$ and this is equivalent to $\frac{1}{6}\leq x\leq \frac{3}{2}$
A: Just posting for enhancement of your knowledge. Algebraic method is not too lengthy. Try to cover up some calculations in mind only. 
$$ \cfrac{2|2x-1|}{|2x+1|} \le 1 \\
\cfrac{|2x-1|}{|2x+1|} \le \cfrac{1}{2} $$
OR I can cross multiply and get : 
$$2|2x-1| \le |2x+1| $$
Square both sides
$$\begin{align}
& 4(2x-1)^2 \le (2x+1)^2 \\
& 4(2x-1)^2 - (2x+1)^2 \le 0\\
& (2(2x-1) - (2x+1))(2(2x-1) + (2x+1) ) \le 0 \\
& (4x - 2 - 2x - 1)(4x - 2 + 2x + 1) \le 0\\
& (2x - 3)(6x - 1) \le 0 \\
& (x - 3/2)(x -1/6) \le 0
\end{align}$$
So, it comes out to be : $\frac{1}{6} \le x \le \frac{3}{2} $
P.S : I still suggest that you just go through this method but prefer to do what others suggested. Just keep this process in your mind and when you need it, just go with it. Squaring modulus functions generally help in future. Good Luck!
A: Simply solve the cases so you can drop the absolute values.


*

*You know that $4x-2$ is negative if $x<\frac12$ and positive otherwise.

*You know that $2x+1$ is negative if $x<-\frac12$ and positive otherwise.


Now, you just have $3$ possibilities:


*

*$x < -\frac12$: In this case, both expressions are negative, so $|4x-2| = -(4x-2)$ and $|2x+1| = -(2x+1)$

*$-\frac12 < x < \frac 12$: In this case, $4x-2<0$ and $2x+1>0$, so $|4x-2|=-(4x-2)$ and $|2x+1|=2x+1$

*$x>\frac12$: Both expressions are positive.


In each case, you can solve a very simple inequality very quickly and see if you get any solutions. Keep in mind though, that if you first say $x<10$, for example, and then get a solution $x>15$, the real answer is that for $x<10$, there are NO solutions.
Also, you still have to see what happens if $x=\pm\frac12$, you get different answers for the two!
A: By the difference of squares identity,
$$(4x-2)^2-(2x+1)^2=(6x-1)(2x-3)$$
and this polynomial takes negative values between the roots.
A: It is actually very quick and simple to sketch a graph--even a rough one, which then easily permits you to eliminate cases:

The blue line is the graph of $y = |4x-2|$, and the red line is the graph of $y = |2x+1|$.  From this we immediately see that the solution set of the given inequality corresponds to the set of $x$-values for which the blue graph is less than or equal to the red.  So we can now solve for the intersection points $4x-2 = 2x+1$ and $-(4x-2) = 2x+1$.
