How to solve inequalities with absolute values on both sides? If you have an inequality that has two absolute value bars like $|4x+1|<|3x|$, how do you go about doing this? I know that if $4x+1<3x$, then those $x$'s will work but what else do I do? I think you do $4x+1<-3x$. Is this correct?
 A: Just take the different cases. For example:
You know that
$$
|3x|=\left\{ \begin{align}
3x & \text{   , if }x\geq 0 \\
-3x & \text{   , if }x <0 
\end{align}
\right\}
$$
$$
|4x+1|=\left\{ \begin{align}
4x+1 & \text{   , if }x\geq \frac{-1}{4} \\
-(4x+1) & \text{   , if }x <\frac{-1}{4}
\end{align}
\right\}
$$
This gives you a few different cases to check: $x<\frac{-1}{4}$, $\frac{-1}{4}\leq x <0$, and $x\geq 0$.
So for instance, take the first cases: $x<\frac{-1}{4}$ so that $|4x+1|=-(4x+1)$ and $|3x|=-3x$. Then we have
$$
\begin{align}
|4x+1|&<|3x| \\
-(4x+1)&<-3x \\
-4x-1&<-3x \\
-x-1&<0 \\
-x&<+1\\
x&>-1
\end{align}
$$
Just be sure to check the $\leq$ or $\geq$ cases and be sure the answers from the different regions agree! (Meaning if you found in one case $x>1$ and found in another case $x<1$ there would be no solutions)
A: There's three parts here that you need to consider:  the area where both 4x+1 and 3x are positive (0 < x); the area where one's positive and the other's negative (-1/4 < x < 0); the area where both are negative (x < -1/4).  Union the results of those together.
A: You could also square everything
$$
|f(x)| < |g(x)| \Leftrightarrow |f(x)|^2 < |g(x)|^2 \\
\Leftrightarrow f(x)^2 < g(x)^2 \\
\Leftrightarrow 0< g(x)^2-f(x)^2 \\
\Leftrightarrow 0< (g(x)-f(x))(g(x)+f(x)), \\
$$
which means that $g(x)-f(x)$ and $g(x)+f(x)$ have the same sign.
A: Need to break up into 2 cases:
(3x) >= 0 or x >= 0:
|3x| = 3x
-3x < (4x+1) < 3x
7x > -1 and x < -1 and x>=0 (no solution)
(3x) <= 0 or x<=0
|3x| = -3x
3x < (4x+1) < -3x
x > -1 and 7x < -1
Solution: -1 < x < -1/7
A: I just solve the equalities without absolute values: LHS = RHS and LHS = -RHS. Then I use a number line test to check values in each interval to determine the intervals of the solution set. This led me to all values between -1 and -1/7, not including engpoints, since the original inequality was strict.
