solve $|2x - 1| <|4x + 1|$ I guess my question is just how to solve those types of inequalities. I know how to solve exercises when i'm presented one absolute value but not two.
 A: Although I can see that this inequality can be solved in inventive ways (e.g. by squaring both sides, as noted in other answers), I would show you the pedestrian way. It is slow and error-prone, but worth considering, because it may apply to other, more complicated situations, where there is no outright "inventive" solution, and also as a reassurance that, as a bottom line, this is a very straightforward (though laborious) problem.
What $|2x-1|$ is depends on whether $2x-1\ge 0$ or $2x-1\lt 0$ (i.e. $x\ge\frac{1}{2}$ or $x\lt\frac{1}{2}$). The same applies to $|4x+1|$: what that is depends on whether $4x+1\ge 0$ or $4x+1\lt 0$, i.e. whether $x\ge-\frac{1}{4}$ or $x\lt-\frac{1}{4}$. So, one way of solving your inequality is: dividing the problems into cases, then solving each case separately, then joining them up together.
Altogether, you would have (up to) $2\times 2=4$ different cases here:


*

*$x\ge\frac{1}{2}$ and $x\ge-\frac{1}{4}$

*$x\ge\frac{1}{2}$ and $x\lt-\frac{1}{4}$

*$x\lt\frac{1}{2}$ and $x\ge-\frac{1}{4}$

*$x\lt\frac{1}{2}$ and $x\lt-\frac{1}{4}$
However, you see immediately that those cases really are:


*

*$x\ge\frac{1}{2}$

*This case cannot happen.

*$-\frac{1}{4}\le x\lt\frac{1}{2}$

*$x\lt-\frac{1}{4}$
Now you would need to solve your inequality in each of the cases 1,3 and 4:


*

*$x\ge\frac{1}{2}$, we have both $2x-1\ge 0$ and $4x+1\ge 0$, so your inequality is reduced to $2x-1\lt 4x+1$, i.e. $x\gt -1$. Obviously, every $x\ge\frac{1}{2}$ satisfies this.

*(ignore)

*$-\frac{1}{4}\le x\lt\frac{1}{2}$: in this case $2x-1\lt 0$ and $4x+1\ge 0$, so your inequality is reduced to $-(2x-1)\lt 4x+1$, i.e. $x\gt 0$. Thus, in this interval the solutions are $0\lt x\lt\frac{1}{2}$.

*$x\lt-\frac{1}{4}$: in this case $2x-1\lt 0$ and $4x+1\lt 0$, so your inequality reduces to $-(2x-1)\lt -(4x+1)$, i.e. $x\lt -1$. Thus, in this interval, all $x\lt -1$ satisfy this inequality.


Now it is time to put these all together and conclude that the overall solution is $x\lt -1$ or $x\gt 0$.
A: If you square it you get (remember that $|a|^2=a^2$) $$4x^2-4x+1<16x^2+8x+1$$
Can you finish it now?
A: Notice that $-\frac14$ is not a solution. So divide on both sides by $|4x+1|$. You are dividing by a positive quantity, so no change in inequality direction. And:
$$\left\lvert\frac{2x-1}{4x+1}\right\rvert<1$$
Now there is just one absolute value. If you prefer:
$$-1<\frac{2x-1}{4x+1}<1$$
If you proceed by multiplying by $4x+1$, you need to distinguish between cases where $4x+1$ is positive and where it is negative. So that you can change inequality directions as needed.
Perhaps better:
$$-1<\frac{\frac{1}{2}(4x+1)-\frac{3}{2}}{4x+1}<1$$
$$-1<\frac{1}{2}-\frac{\frac{3}{2}}{4x+1}<1$$
$$-\frac{3}{2}<-\frac{\frac{3}{2}}{4x+1}<\frac{1}{2}$$
$$1>\frac{1}{4x+1}>-\frac{1}{3}$$
And proceed from here, again being careful with when the inequality direction changes.
A: Hint
If $a>0$ then $$|y|<a \Leftrightarrow -a<y<a$$ and $$|y|>a \Leftrightarrow y<-a \text{ or } y>a$$
