Answer:
The 1st and the 5th tables represent the same function
Step-by-step explanation:
* Lets explain how to solve the problem
- There are five tables of functions, two of them are equal
- To find the two equal function lets find their equations
- The form of the equation of a line whose endpoints are (x1 , y1) and
(x2 , y2) is 
* Lets make the equation of each table
# (x1 , y1) = (4 , 8) and (x2 , y2) = (6 , 7)
∵ x1 = 4 , x2 = 6 and y1 = 8 , y2 = 7
∴ 
∴ 
- By using cross multiplication
∴ 2(y - 8) = -1(x - 4) ⇒ simplify
∴ 2y - 16 = -x + 4 ⇒ add x and 16 for two sides
∴ x + 2y = 20 ⇒ (1)
# (x1 , y1) = (4 , 5) and (x2 , y2) = (6 , 4)
∵ x1 = 4 , x2 = 6 and y1 = 5 , y2 = 4
∴ 
∴ 
- By using cross multiplication
∴ 2(y - 5) = -1(x - 4) ⇒ simplify
∴ 2y - 10 = -x + 4 ⇒ add x and 10 for two sides
∴ x + 2y = 14 ⇒ (2)
# (x1 , y1) = (2 , 8) and (x2 , y2) = (8 , 5)
∵ x1 = 2 , x2 = 8 and y1 = 8 , y2 = 5
∴ 
∴ 
- By using cross multiplication
∴ 2(y - 8) = -1(x - 2) ⇒ simplify
∴ 2y - 16 = -x + 2 ⇒ add x and 16 for two sides
∴ x + 2y = 18 ⇒ (3)
# (x1 , y1) = (2 , 10) and (x2 , y2) = (6 , 14)
∵ x1 = 2 , x2 = 6 and y1 = 10 , y2 = 14
∴ 
∴ 
- By using cross multiplication
∴ (y - 10) = (x - 2)
∴ y - 10 = x - 2 ⇒ add 2 and subtract y in the two sides
∴ -8 = x - y ⇒ switch the two sides
∴ x - y = -8 ⇒ (4)
# (x1 , y1) = (2 , 9) and (x2 , y2) = (8 , 6)
∵ x1 = 2 , x2 = 8 and y1 = 9 , y2 = 6
∴ 
∴ 
- By using cross multiplication
∴ 2(y - 9) = -1(x - 2) ⇒ simplify
∴ 2y - 18 = -x + 2 ⇒ add x and 18 for two sides
∴ x + 2y = 20 ⇒ (5)
- Equations (1) and (5) are the same
∴ The 1st and the 5th tables represent the same function