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.
$\begin{array}{c|c} x & y \\ \hline 2 & 5 \\ p & -14 \\ -2 & -17 \\ \end{array}$
+874 Slope: $\frac{5-(-17)}{2-(-2)} = \frac{22}{4} = \frac{11}{4}$
We have: $y = \frac{11}{4} x + b$
Substitute one value of $(x,y)$ to find $b:$ $5 = \frac{11}{4} \cdot 2 + b$
$b= -\frac{1}{2}$
$y = \frac{11}{4} x - \frac{1}{2}$
$-14 = \frac{11}{4} p - \frac{1}{2}$
$- \frac{27}{2} = \frac{11}{4} p$
$- 54 = 11p$
$p = - \frac{54}{11}$
$q = \frac{11}{4} \cdot 13 - \frac{1}{2}$
$q = \frac{141}{4}$
$p + q = - \frac{54}{11} + \frac{141}{4} \approx \boxed{30.3}$
+874 Slope: $\frac{5-(-17)}{2-(-2)} = \frac{22}{4} = \frac{11}{4}$
We have: $y = \frac{11}{4} x + b$
Substitute one value of $(x,y)$ to find $b:$ $5 = \frac{11}{4} \cdot 2 + b$
$b= -\frac{1}{2}$
$y = \frac{11}{4} x - \frac{1}{2}$
$-14 = \frac{11}{4} p - \frac{1}{2}$
$- \frac{27}{2} = \frac{11}{4} p$
$- 54 = 11p$
$p = - \frac{54}{11}$
$q = \frac{11}{4} \cdot 13 - \frac{1}{2}$
$q = \frac{141}{4}$
$p + q = - \frac{54}{11} + \frac{141}{4} \approx \boxed{30.3}$
