I find in article ,,Model-Based Recognition of 3D Objects from One View,, by I. Weiss M. Ray this system of equations:
$$\mu m'_{12} - I_1\mu'm_{12} + \mu\mu'(I_1 - 1)m_{25} = 0 \\ \mu m'_{13} - I_2\mu'm_{13} + \mu\mu'(I_2 - 1)m_{35} = 0 \\ \mu m'_{14} - I_3\mu'm_{14} + \mu\mu'(I_3 - 1)m_{45} = 0 \\ $$
And somehow eliminating $\mu$ and $\mu'$ you get:
$$ I_3(I_2 - 1)i_1i_2 - I_3(I_1 - 1)i_1 - I_1(I_2 - 1)i_2 = I_2(I_3 - 1)i_3i_4 - I_2(I_1 - 1)i_3 - I_1(I_3 - 1)i_4$$ where: $i_1 = \frac{m'_{12}m_{14}}{m_{12}m'_{14}}$ $i_2 = \frac{m'_{12}m_{35}}{m'_{13}m_{25}}$ $i_3 = \frac{m'_{12}m_{13}}{m_{12}m'_{13}}$ $i_4 = \frac{m'_{12}m_{45}}{m'_{14}m_{25}}$
There's also comment:
It is easy to elimnate the terms proportional to $\mu$, $\mu'$. The remaining (last above) equation is devided by $\mu\mu'm_{12}m'_{13}m'_{14}m_{25}$
I would like to know how to eliminate $\mu$ and $\mu'$ step by step. Any solution is good, even if final equation differ.
Divide the above equations by $\mu\mu'$, and you get a linear system in $\dfrac1\mu$ and $\dfrac1{\mu'}$, on which you can operate by row reduction.