problem: The points $(x, y)$ represented in this table lie on a straight line. The point $(13, q)$ lies on the same line. What is the value of $p + q?$ Express your answer as a decimal to the nearest tenth. $$\begin{array}{c|c}
  x & y \\ \hline
  2 & -5 \\
  p & -14 \\
  p+2 & -17 \\
  \end{array}$$
solution: If we have two points $(x_1,y_1)$ and $(x_2,y_2)$ on a line, we can find the slope of the line by using the formula $\dfrac{y_1-y_2}{x_1-x_2}.$ So, for the line we are given, the slope is $\dfrac{(-5)-(-14)}{2-p}=\dfrac{9}{2-p},$ and the slope is also $\dfrac{(-14)-(-17)}{p-(p+2)}=\dfrac{3}{-2}.$ Setting these values equal, we get $$\dfrac{9}{2-p}=-\dfrac{3}{2}.$$ Multiplying both sides by the product of the denominators and simplifying gives \begin{align*}
  (2-p)(3)&=(-2)(9)\\
  6-3p&=-18 \\
  p&=8.
  \end{align*} Now we need to find $q.$ Using the same strategy as above, we find that \begin{align*}
  \frac{q-(-5)}{13-2}&=\frac{3}{-2} \\
  (11)(3)&=(-2)(q+5)\\
  33&=-2q-10 \\
  q&=-21.5.\\
  \end{align*} Therefore, $p+q=8+(-21.5)=\boxed{-13.5}.$