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High School Math Algebra High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

How many of each type of vehicle could transport all 31 people to the ski area in one trip. The number of van are 3 and number of sedan are 2. Given information: It is given that there are 26 members in a group and five chaperons. The number of seat in van is7 and number of seats in sedan are five. Concept used : Substitution. Method of solving system of linear equation Calculation : Let the number of van be x and number of sedan be y . There are 5 chaperons who drive either van or sedan that means total number of vehicles are 5. Therefore, it can be written as: x + y = 5 In van 7 people can sit so total persons in van is 7 x . In sedan total five people can sit so total person in sedan is 5 y Total person that either sit in van or sedan are 31. Hence equation obtained 7 x + 5 y = 31 . Thus, the system of equation is x + y = 5 .... ( 1 ) 7 x + 5 y = 31... ( 2 ) Multiply equation (1) by 7 it becomes: 7 x + 7 y = 35... ( 3 ) Now subtract equation (2) from (3): 7 x − 7 x + 7 y − 5 y = 35 − 31 0 + 2 y = 4 2 y = 4 y = 4 2 y = 2 Substitute the value of y in equation (1): x + 2 = 5 x = 5 − 2 x = 3 The number of van are 3 and number of sedan are 2.

How many of each type of vehicle could transport all 31 people to the ski area in one trip. The number of van are 3 and number of sedan are 2. Given information: It is given that there are 26 members in a group and five chaperons. The number of seat in van is7 and number of seats in sedan are five. Concept used : Substitution. Method of solving system of linear equation Calculation : Let the number of van be x and number of sedan be y . There are 5 chaperons who drive either van or sedan that means total number of vehicles are 5. Therefore, it can be written as: x + y = 5 In van 7 people can sit so total persons in van is 7 x . In sedan total five people can sit so total person in sedan is 5 y Total person that either sit in van or sedan are 31. Hence equation obtained 7 x + 5 y = 31 . Thus, the system of equation is x + y = 5 .... ( 1 ) 7 x + 5 y = 31... ( 2 ) Multiply equation (1) by 7 it becomes: 7 x + 7 y = 35... ( 3 ) Now subtract equation (2) from (3): 7 x − 7 x + 7 y − 5 y = 35 − 31 0 + 2 y = 4 2 y = 4 y = 4 2 y = 2 Substitute the value of y in equation (1): x + 2 = 5 x = 5 − 2 x = 3 The number of van are 3 and number of sedan are 2.

Solution Summary: The author calculates how many of each type of vehicle could transport all 31 people to the ski area in one trip.

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

15th Edition

ISBN: 9780133281163

Author: Prentice Hall

Publisher: Prentice Hall

See similar textbooks

Expert Solution & Answer

Chapter 3.2, Problem 21PPSE

Solution

To find: How many of each type of vehicle could transport all 31 people to the ski area in one trip.

The number of van are 3 and number of sedan are 2.

Given information:

It is given that there are 26 members in a group and five chaperons. The number of seat in van is7 and number of seats in sedan are five.

Concept used: Substitution. Method of solving system of linear equation

Calculation:

Let the number of van be x and number of sedan be y .

There are 5 chaperons who drive either van or sedan that means total number of vehicles are 5.

Therefore, it can be written as:

x+y=5

In van 7 people can sit so total persons in van is 7x .

In sedan total five people can sit so total person in sedan is 5y

Total person that either sit in van or sedan are 31.

Hence equation obtained

7x+5y=31 .

Thus, the system of equation is

x+y=5 ....(1)7x+5y=31...(2)

Multiply equation (1) by 7 it becomes:

7x+7y=35...(3)

Now subtract equation (2) from (3):

7x−7x+7y−5y=35−31 0+2y=4 2y=4 y=42 y=2

Substitute the value of y in equation (1):

x+2=5 x=5−2 x=3

The number of van are 3 and number of sedan are 2.

Chapter 3.2, Problem 20PPSE

Chapter 3.2, Problem 22PPSE

Chapter 3 Solutions

High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11

Ch. 3.1 - Prob. 1LC Ch. 3.1 - Prob. 2LC Ch. 3.1 - Prob. 3LC Ch. 3.1 - Prob. 4LC Ch. 3.1 - Prob. 5LC Ch. 3.1 - Prob. 6LC Ch. 3.1 - Prob. 7PPSE Ch. 3.1 - Prob. 8PPSE Ch. 3.1 - Prob. 9PPSE Ch. 3.1 - Prob. 10PPSE

Ch. 3.1 - Prob. 11PPSE Ch. 3.1 - Prob. 12PPSE Ch. 3.1 - Prob. 13PPSE Ch. 3.1 - Prob. 14PPSE Ch. 3.1 - Prob. 15PPSE Ch. 3.1 - Prob. 16PPSE Ch. 3.1 - Prob. 17PPSE Ch. 3.1 - Prob. 18PPSE Ch. 3.1 - Prob. 19PPSE Ch. 3.1 - Prob. 20PPSE Ch. 3.1 - Prob. 21PPSE Ch. 3.1 - Prob. 22PPSE Ch. 3.1 - Prob. 23PPSE Ch. 3.1 - Prob. 24PPSE Ch. 3.1 - Prob. 25PPSE Ch. 3.1 - Prob. 26PPSE Ch. 3.1 - Prob. 27PPSE Ch. 3.1 - Prob. 28PPSE Ch. 3.1 - Prob. 29PPSE Ch. 3.1 - Prob. 30PPSE Ch. 3.1 - Prob. 31PPSE Ch. 3.1 - Prob. 32PPSE Ch. 3.1 - Prob. 33PPSE Ch. 3.1 - Prob. 34PPSE Ch. 3.1 - Prob. 35PPSE Ch. 3.1 - Prob. 36PPSE Ch. 3.1 - Prob. 37PPSE Ch. 3.1 - Prob. 38PPSE Ch. 3.1 - Prob. 39PPSE Ch. 3.1 - Prob. 40PPSE Ch. 3.1 - Prob. 41PPSE Ch. 3.1 - Prob. 42PPSE Ch. 3.1 - Prob. 43PPSE Ch. 3.1 - Prob. 44PPSE Ch. 3.1 - Prob. 45PPSE Ch. 3.1 - Prob. 46PPSE Ch. 3.1 - Prob. 47PPSE Ch. 3.1 - Prob. 48PPSE Ch. 3.1 - Prob. 49PPSE Ch. 3.1 - Prob. 50PPSE Ch. 3.1 - Prob. 51PPSE Ch. 3.1 - Prob. 52PPSE Ch. 3.1 - Prob. 53PPSE Ch. 3.1 - Prob. 54PPSE Ch. 3.1 - 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