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 ![\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}](https://tex.z-dn.net/?f=%5Cfrac%7By-y_%7B1%7D%7D%7Bx-x_%7B1%7D%7D%3D%5Cfrac%7By_%7B2%7D-y_%7B1%7D%7D%7Bx_%7B2%7D-x_%7B1%7D%7D)
* 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
∴ ![\frac{y-8}{x-4}=\frac{7-8}{6-4}](https://tex.z-dn.net/?f=%5Cfrac%7By-8%7D%7Bx-4%7D%3D%5Cfrac%7B7-8%7D%7B6-4%7D)
∴ ![\frac{y-8}{x-4}=\frac{-1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7By-8%7D%7Bx-4%7D%3D%5Cfrac%7B-1%7D%7B2%7D)
- 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
∴ ![\frac{y-5}{x-4}=\frac{4-5}{6-4}](https://tex.z-dn.net/?f=%5Cfrac%7By-5%7D%7Bx-4%7D%3D%5Cfrac%7B4-5%7D%7B6-4%7D)
∴ ![\frac{y-5}{x-4}=\frac{-1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7By-5%7D%7Bx-4%7D%3D%5Cfrac%7B-1%7D%7B2%7D)
- 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
∴ ![\frac{y-8}{x-2}=\frac{5-8}{8-2}](https://tex.z-dn.net/?f=%5Cfrac%7By-8%7D%7Bx-2%7D%3D%5Cfrac%7B5-8%7D%7B8-2%7D)
∴ ![\frac{y-8}{x-2}=\frac{-3}{6}=====\frac{y-8}{x-2}=\frac{-1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7By-8%7D%7Bx-2%7D%3D%5Cfrac%7B-3%7D%7B6%7D%3D%3D%3D%3D%3D%5Cfrac%7By-8%7D%7Bx-2%7D%3D%5Cfrac%7B-1%7D%7B2%7D)
- 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
∴ ![\frac{y-10}{x-2}=\frac{14-10}{6-2}](https://tex.z-dn.net/?f=%5Cfrac%7By-10%7D%7Bx-2%7D%3D%5Cfrac%7B14-10%7D%7B6-2%7D)
∴ ![\frac{y-10}{x-2}=\frac{4}{4}======\frac{y-10}{x-2}=1](https://tex.z-dn.net/?f=%5Cfrac%7By-10%7D%7Bx-2%7D%3D%5Cfrac%7B4%7D%7B4%7D%3D%3D%3D%3D%3D%3D%5Cfrac%7By-10%7D%7Bx-2%7D%3D1)
- 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
∴ ![\frac{y-9}{x-2}=\frac{6-9}{8-2}](https://tex.z-dn.net/?f=%5Cfrac%7By-9%7D%7Bx-2%7D%3D%5Cfrac%7B6-9%7D%7B8-2%7D)
∴ ![\frac{y-9}{x-2}=\frac{-3}{6}======\frac{y-9}{x-2}=\frac{-1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7By-9%7D%7Bx-2%7D%3D%5Cfrac%7B-3%7D%7B6%7D%3D%3D%3D%3D%3D%3D%5Cfrac%7By-9%7D%7Bx-2%7D%3D%5Cfrac%7B-1%7D%7B2%7D)
- 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