[Geometry]
Problem: Determine the coordinates of the points of intersection of the graphs of $y=\log _{10}(x-2)$ and $y=1-\log _{10}(x+1)$.

Solution: The intersection takes place where,

$$
\begin{aligned}
& \log _{10}(x-2)=1-\log _{10}(x+1) \\
& \log _{10}(x-2)+\log _{10}(x+1)=1 \\
& \log _{10}\left(x^{2}-x-2\right)=1
\end{aligned}
$$



$$
\begin{aligned}
& x^{2}-x-2=10 \\
& x^{2}-x-12=0 \\
& (x-4)(x+3)=0 \\
& x=4 \text { or }-3
\end{aligned}
$$

For $x=-3, y$ is not defined.

For $x=4, y=\log _{10} 2 \doteq 0.3$.

The graphs therefore intersect at $\left(4, \log _{10} 2\right)$.

Final answer: $(4, \log _{10} 2)$