problem: Let $P=(a,b)$ be the point of intersection of the line $y=2x-10$ and the line through $(7,8)$ and $(9,0)$.  Compute $a+b$.
solution: The slope of the line through $(7,8)$ and $(9,0)$ is $\frac{8-0}{7-9}=\frac{8}{-2}=-4$.  Thus, the line has equation $y=-4x+b$ for some $b$.  Since $B(9,0)$ lies on this line, we have $0=-4(9)+b \Rightarrow b=36$, and thus the equation of the line is $y=-4x+36$.
  
  To determine the point of intersection between the lines having equations $y=-4x+36$ and $y=2x-10$, we set the two expressions for $y$ equal to each other and solve for $x$.  We have $-4x+36=2x-10 \Rightarrow x = \frac{23}{3}$.  It follows that  $y=2x-10=2\left(\frac{23}{3}\right)-10 = \frac{46}{3}-\frac{30}{3}=\frac{16}{3}$.
  
  Thus, $P=(\frac{23}{3},\frac{16}{3})$ and $a+b=\frac{23}{3}+\frac{16}{3}=\frac{39}{3}=\boxed{13}$.