(Should be) simple algebra exercise I have the following coupled equations
\begin{equation}
\tag{1}
a_{1}\left(T-T_{m}\right)+b_{1}M^{2}+cP^{2}=0
\end{equation}
\begin{equation}
\tag{2}
a_{2}\left(T-\theta_{2}\right)+b_{2}P^{2}+cM^{2}=0
\end{equation}
\begin{equation}
\tag{3}
\theta_{2}=\frac{ca_{1}}{a_{2}b_{1}}\left(T_{m}-T_{e}\right)+T_{e}
\end{equation}
I have to show that
$$
M^{2}=\frac{a_{1}}{b_{1}}\left(T_{m}-T_{e}\right)+\frac{a_{1}b_{2}-a_{2}c}{b_{1}b_{2}-c^{2}}\left(T_{e}-T\right)
$$
Attempt:
Doing $b_{2}\left(1\right)-c\left(2\right)$ I get to
$$
M^{2}\left(b_{1}b_{2}-c^{2}\right)=a_{2}c\left(T-T_{e}\right)+a_{1}b_{2}\left(T_{m}-T\right)+\frac{a_{1}c^{2}}{b_{1}}\left(T_{e}-T_{m}\right)
$$
How to finish?


$\displaystyle
M^{2}\left(b_{1}b_{2}-c^{2}\right)=a_{2}c\left(T-T_{e}\right)+a_{1}b_{2}\left(T_{m}-T\right)+\frac{a_{1}c^{2}}{b_{1}}\left(T_{e}-T_{m}\right)
$

$$
\begin{align}
M^{2}\left(b_{1}b_{2}-c^{2}\right) &= a_{2}c\left(T-T_{e}\right)+a_{1}b_{2}\left(T_{m} \color{red}{-T_e+T_e} -T\right)+\frac{a_{1}c^{2}}{b_{1}}\left(T_{e}-T_{m}\right) \\
 &= \left(a_1b_2-\frac{a_1c^2}{b_1}\right)(T_m-T_e)+\left(-a_2c+a_1b_2\right)(T_e-T) \\
 &= \frac{a_1}{b_1}(b_1b_2-c^2)(T_m-T_e) + +\left(a_1b_2-a_2c\right)(T_e-T)
\end{align}
$$
