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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.

 

$\begin{array}{c|c} x & y \\ \hline 2 & 5 \\ p & -14 \\ -2 & -17 \\ \end{array}$

 Jun 27, 2021

Best Answer 

 #1
avatar+785 
+1

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 $b:$ $5 = \frac{11}{4} \cdot 2 + b$

$b= -\frac{1}{2}$

$y = \frac{11}{4} x - \frac{1}{2}$

$-14 = \frac{11}{4} p - \frac{1}{2}$

$- \frac{27}{2} = \frac{11}{4} p$

$- 54 = 11p$

$p = - \frac{54}{11}$

$q = \frac{11}{4} \cdot 13 - \frac{1}{2}$

$q = \frac{141}{4}$

$p + q = - \frac{54}{11} + \frac{141}{4} \approx \boxed{30.3}$

 Jun 27, 2021
 #1
avatar+785 
+1
Best Answer

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 $b:$ $5 = \frac{11}{4} \cdot 2 + b$

$b= -\frac{1}{2}$

$y = \frac{11}{4} x - \frac{1}{2}$

$-14 = \frac{11}{4} p - \frac{1}{2}$

$- \frac{27}{2} = \frac{11}{4} p$

$- 54 = 11p$

$p = - \frac{54}{11}$

$q = \frac{11}{4} \cdot 13 - \frac{1}{2}$

$q = \frac{141}{4}$

$p + q = - \frac{54}{11} + \frac{141}{4} \approx \boxed{30.3}$

MathProblemSolver101 Jun 27, 2021

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