On page 272 of Banchoff and Lovett's book "Differential Geometry of Curves and Surfaces," they make the following argument (paraphrasing):
For a parameterization of a surface embedded in $\mathbb{R}^3$, $\vec{X}$, a curve parameterized by arclength, $\gamma=X\circ\alpha$ (where $\alpha(s)=(u(s),v(s))^T$) is a geodesic only if \begin{equation} \frac{d^2x^i}{ds^2} + \Gamma_{jk}^i\frac{dx^j}{ds}\frac{dx^k}{ds}=0. (i=1,2; x=(u(s),v(s))). \end{equation} They arrive at this by noting that, since the derivative of the tangent vector to a geodesic lies strictly along the normal (since geodesic curvature must be zero), $\vec{T}'\cdot \vec{X}_u=\vec{T}'\cdot\vec{X}_v=0$ (where $T$ is the tangent vector to $\gamma$), and so \begin{align} \vec{X}_{uu}\cdot \vec{X}_u(u')^2+2\vec{X}_{uv}\cdot\vec{X}_uu'v'+\vec{X}_{vv}\cdot{X}_u(v')^2+\vec{X}_u\cdot\vec{X}_uu''+\vec{X}_v\cdot\vec{X}_uv''&=0\\ \vec{X}_{uu}\cdot\vec{X}_v(u')^2+2\vec{X}_{uv}\cdot\vec{X}_vu'v'+\vec{X}_{vv}\cdot{X}_v(v')^2+\vec{X}_u\cdot\vec{X}_vu''+\vec{X}_v\cdot\vec{X}_vv''&=0. \end{align} I notice that the first three terms in each of these equations is the Christoffel symbol term from the equation above and that $d^2x^i/ds^2$ is the $u''$ or $v''$ term, but I can't for the life of me understand how the $v''$ ($u''$) term drops away in the first (second) equation. How is $\vec{X}_u\cdot\vec{X}_uu''+\vec{X}_v\cdot\vec{X}_uv''=u''$ and $\vec{X}_u\cdot\vec{X}_vu''+\vec{X}_u\cdot\vec{X}_vv''=v''$?
Lovett and Banchoff say this after listing the two long equations: "Solving algebraically for $u''$ and $v''$ in the above two equations, we obtain the following classical equations for a geodesic curve which we will express in tensor notation for simplicitly:" (followed by the equation I wrote in the beginning). This makes me think that there is some algebraic manipulation to prove it, but I can't figure out what it is. Any help would be greatly appreciated!