MHB Lagrangian Mechanics: Solving $\mathcal{L}(X,x,\dot{X},\dot{x})$ for 2 Masses

[Dustinsfl]

Write down the Lagrangian $\mathcal{L}(x_1,x_2,\dot{x}_1,\dot{x}_2)$ for two particles of equal masses, $m_1 = m_2 = m$, confined to the $x$ axis and connected by a spring with potential energy $U = \frac{1}{2}kx^2$. [Here $x$ is the extension of the spring, $x = (x_1 - x_2 - \ell)$ where $\ell$ is the spring's unstretched length, and I assume that mass 1 remains to the right of mass 2 at all times.]

The Lagrangian, $\mathcal{L} = T - U$, is the kinetic minus the potential energy.
\begin{alignat*}{3}
\mathcal{L} & = & T - U\\
& = & \frac{1}{2}m\left(\dot{x}_1^2 + \dot{x}_2^2\right) - \frac{1}{2}k(x_1 - x_2 - \ell)^2
\end{alignat*}

Rewrite $\mathcal{L}$ in terms of the new variables $X = \frac{1}{2}(x_1 + x_2)$ (the CM position) and $x$ (the extension), and write down the two Lagrange equations for $X$ and $x$.

The solution has:
Let $x = x_1 - x_2 - \ell$
Where did this piece come from (below). I see that adding them together produces $X$.
\begin{alignat}{3}
x_1 & = & X + \frac{x}{2} + \frac{\ell}{2}\\
x_2 & = & X - \frac{x}{2} - \frac{\ell}{2}
\end{alignat}

 

[Ackbach]

You have two equations:
\begin{align*}
X&= \frac{1}{2}\,(x_{1}+x_{2})\\
x&= x_{1}-x_{2}-\ell.
\end{align*}
In order to see how $L$ is using these new variables, you must solve for the original variables $x_{1}$ and $x_{2}$, and then plug those expressions into the $L$ that you know. So the two equations at the bottom of your post are simply using standard solving methods (elimination, substitution) to write the old variables in terms of the new.