$$\left | \frac{4-3m_3}{3+4m_3} \right |= \left | \frac{-3-4m_3}{4-3m_3} \right |$$
I am always confused when it comes to modulus. I know if there is modulus any one of the side then when we remove it we get two cases($\pm$) just like when we remove square.
I know normally S.E. doesn't help with simplification, bu i am really stuck here in my problem.
Dropping the subsript to make typing easier, and assuming $m$ is real we have $(4-3m)^2=(3+4m)^2$, so $7m^2+48m-7=0$. Factorising $(7m-1)(m+7)=0$, so $m=1/7$ or -7.
With a little more effort you can get the additional solutions $m=\pm i$ if you are allowing complex numbers.