Algebra and Trigonometry: Structure and Method, Book 2
Algebra and Trigonometry: Structure and Method, Book 2
2000th Edition
ISBN: 9780395977255
Author: MCDOUGAL LITTEL
Publisher: McDougal Littell
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Concept explainers

Unitary Method
The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.
Speed, Time, and Distance
Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 …
Profit and Loss
The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.
Units and Measurements
Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experime…
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Question
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Chapter 3.6, Problem 2P
To determine

To find:the number of single tickets and couple tickets sold.

Expert Solution & Answer
Check Mark

Answer to Problem 2P

The number of single tickets and couple tickets sold are 16 and 56, respectively.

Explanation of Solution

Given information:

The cost of each single ticket and each couple ticketare $20 and $35, respectively.

The total cost of ticket sales is $2280 and total number of attended people is 128.

Calculation:

Let x be a number of single tickets sold.

Let y be a number of couple tickets sold.

The total number of attended people is 128, so that x+2y=128   ....... (1)

The total cost of ticket sales is $2280 and the cost of each single tickets and couple tickets are $20 and $35, respectively.

It can be converted into linear equation as 20x+35y=2280   ....... (2)

Compute the solutions of the system x+2y=128 and 20x+35y=2280 by linear combination method.

Find the linear combination that eliminates the variable y .

Multiply the first equation by −20 . that is, −20x−40y=−2560   .......(3)

Add the equation (2) and (3).

  −20x−40y=−2560  20x+35y=22800x−5y=−280¯

Divide on both sides by −5 ,

  y=−280−5y=56

Substitute y=56 in x+2y=128 and simplify the value of x ,

  x+2(56)=128x+112=128x=128−112x=16

Therefore, the number of single tickets and couple tickets sold are 16 and 56, respectively.

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Chapter 3.6, Problem 1P
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Chapter 3.6, Problem 3P

Chapter 3 Solutions

Algebra and Trigonometry: Structure and Method, Book 2

Ch. 3.1 - Prob. 1OECh. 3.1 - Prob. 2OECh. 3.1 - Prob. 3OECh. 3.1 - Prob. 4OECh. 3.1 - Prob. 5OECh. 3.1 - Prob. 6OECh. 3.1 - Prob. 7OECh. 3.1 - Prob. 8OECh. 3.1 - Prob. 9OECh. 3.1 - Prob. 10OE
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Prob. 2MRECh. 3.9 - Prob. 3MRECh. 3.9 - Prob. 4MRECh. 3.9 - Prob. 5MRECh. 3.9 - Prob. 6MRECh. 3.9 - Prob. 7MRECh. 3.9 - Prob. 8MRECh. 3.9 - Prob. 9MRECh. 3.9 - Prob. 10MRECh. 3.9 - Prob. 11MRECh. 3.9 - Prob. 12MRECh. 3.9 - Prob. 13MRECh. 3.10 - Prob. 1OECh. 3.10 - Prob. 2OECh. 3.10 - Prob. 3OECh. 3.10 - Prob. 4OECh. 3.10 - Prob. 5OECh. 3.10 - Prob. 6OECh. 3.10 - Prob. 7OECh. 3.10 - Prob. 8OECh. 3.10 - Prob. 9OECh. 3.10 - Prob. 10OECh. 3.10 - Prob. 11OECh. 3.10 - Prob. 12OECh. 3.10 - Prob. 13OECh. 3.10 - Prob. 14OECh. 3.10 - Prob. 15OECh. 3.10 - Prob. 16OECh. 3.10 - Prob. 17OECh. 3.10 - Prob. 18OECh. 3.10 - Prob. 19OECh. 3.10 - Prob. 20OECh. 3.10 - Prob. 21OECh. 3.10 - Prob. 22OECh. 3.10 - Prob. 23OECh. 3.10 - Prob. 24OECh. 3.10 - Prob. 1WECh. 3.10 - Prob. 2WECh. 3.10 - Prob. 3WECh. 3.10 - Prob. 4WECh. 3.10 - Prob. 5WECh. 3.10 - Prob. 6WECh. 3.10 - Prob. 7WECh. 3.10 - Prob. 8WECh. 3.10 - Prob. 9WECh. 3.10 - Prob. 10WECh. 3.10 - Prob. 11WECh. 3.10 - Prob. 12WECh. 3.10 - Prob. 13WECh. 3.10 - Prob. 14WECh. 3.10 - Prob. 15WECh. 3.10 - Prob. 16WECh. 3.10 - Prob. 17WECh. 3.10 - Prob. 18WECh. 3.10 - Prob. 19WECh. 3.10 - Prob. 20WECh. 3.10 - Prob. 21WECh. 3.10 - Prob. 22WECh. 3.10 - Prob. 23WECh. 3.10 - Prob. 24WECh. 3.10 - Prob. 25WECh. 3.10 - Prob. 26WECh. 3.10 - Prob. 27WECh. 3.10 - Prob. 28WECh. 3.10 - Prob. 29WECh. 3.10 - Prob. 30WECh. 3.10 - Prob. 31WECh. 3.10 - Prob. 32WECh. 3.10 - Prob. 33WECh. 3.10 - Prob. 34WECh. 3.10 - Prob. 35WECh. 3.10 - Prob. 36WECh. 3.10 - Prob. 37WECh. 3.10 - Prob. 1STCh. 3.10 - Prob. 2STCh. 3.10 - Prob. 3STCh. 3.10 - Prob. 4STCh. 3.10 - Prob. 5STCh. 3.10 - Prob. 6STCh. 3.10 - Prob. 7STCh. 3.10 - Prob. 8STCh. 3.10 - Prob. 9STCh. 3.10 - Prob. 1ECh. 3.10 - Prob. 2ECh. 3.10 - Prob. 3ECh. 3.10 - Prob. 4ECh. 3.10 - Prob. 5ECh. 3 - Prob. 1CRCh. 3 - Prob. 2CRCh. 3 - Prob. 3CRCh. 3 - Prob. 4CRCh. 3 - Prob. 5CRCh. 3 - Prob. 6CRCh. 3 - Prob. 7CRCh. 3 - Prob. 8CRCh. 3 - Prob. 9CRCh. 3 - Prob. 10CRCh. 3 - Prob. 11CRCh. 3 - Prob. 12CRCh. 3 - Prob. 13CRCh. 3 - Prob. 14CRCh. 3 - Prob. 15CRCh. 3 - Prob. 16CRCh. 3 - Prob. 17CRCh. 3 - Prob. 18CRCh. 3 - Prob. 1CTCh. 3 - Prob. 2CTCh. 3 - Prob. 3CTCh. 3 - Prob. 4CTCh. 3 - Prob. 5CTCh. 3 - Prob. 6CTCh. 3 - Prob. 7CTCh. 3 - Prob. 8CTCh. 3 - Prob. 9CTCh. 3 - Prob. 10CTCh. 3 - Prob. 11CTCh. 3 - Prob. 12CTCh. 3 - Prob. 13CTCh. 3 - Prob. 14CTCh. 3 - Prob. 15CTCh. 3 - Prob. 16CTCh. 3 - Prob. 1CLRCh. 3 - Prob. 2CLRCh. 3 - Prob. 3CLRCh. 3 - Prob. 4CLRCh. 3 - Prob. 5CLRCh. 3 - Prob. 6CLRCh. 3 - Prob. 7CLRCh. 3 - Prob. 8CLRCh. 3 - Prob. 9CLRCh. 3 - Prob. 10CLRCh. 3 - Prob. 11CLRCh. 3 - Prob. 12CLRCh. 3 - Prob. 13CLRCh. 3 - Prob. 14CLRCh. 3 - Prob. 15CLRCh. 3 - Prob. 16CLRCh. 3 - Prob. 17CLRCh. 3 - Prob. 18CLRCh. 3 - Prob. 19CLRCh. 3 - Prob. 20CLRCh. 3 - Prob. 21CLRCh. 3 - Prob. 22CLRCh. 3 - Prob. 23CLRCh. 3 - Prob. 24CLRCh. 3 - Prob. 25CLRCh. 3 - Prob. 26CLRCh. 3 - Prob. 27CLRCh. 3 - Prob. 28CLR

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