$$\left | \frac{4-3m_3}{3+4m_3} \right |= \left | \frac{-3-4m_3}{4-3m_3} \right |$$

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.

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    $\begingroup$ There's an $m_2$ on the left side, but only $m_3$s elsewhere. Should these all just be the single unknown $m$? Or is there some typographical error? $\endgroup$ – John Hughes Sep 3 '14 at 16:19
  • $\begingroup$ Are the $m_2,m_3$ real? $\endgroup$ – almagest Sep 3 '14 at 16:20
  • $\begingroup$ @almagest I have edited it's only $m_3$ $\endgroup$ – RutvikSutaria Sep 3 '14 at 16:21

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.

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  • $\begingroup$ Sorry, that was my typo, the factor is $m+7$. The solutions are -7 and 1/7. $\endgroup$ – almagest Sep 3 '14 at 16:32

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