(Should be) simple algebra exercise

Question

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?

Answer 1

$\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) $

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