How to solve inequalities with absolute values on both sides? If you have an inequality that has two absolute value bars like $|4x+1|&lt;|3x|$, how do you go about doing this? I know that if $4x+1&lt;3x$, then those $x$'s will work but what else do I do? I think you do $4x+1&lt;-3x$. Is this correct?
You could also square everything
$$
|f(x)| &lt; |g(x)| \Leftrightarrow |f(x)|^2 &lt; |g(x)|^2 \\
\Leftrightarrow f(x)^2 &lt; g(x)^2 \\
\Leftrightarrow 0&lt; g(x)^2-f(x)^2 \\
\Leftrightarrow 0&lt; (g(x)-f(x))(g(x)+f(x)), \\
$$
which means that $g(x)-f(x)$ and $g(x)+f(x)$ have the same sign.
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.