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

How many more dimes than quarters does Georgia have in the given situation. Georgia has 2 dimes more than quarters. Given: Georgia has only dimes and quarters. She has total 18 coins that are worth $ 3 . Concepts Used: Modelling a situational problem mathematically. Solving a pair of linear equations in two variables using elimination method. Calculations: Let number of dimes be x and the number of quarters be y . The total number of coins is given to be 18 . Represent this mathematically: x + y = 18 . Recall that 1 dime is worth $ 0.10 . It follows that x dimes are worth 0.1 x dollars. Also recall that 1 quarter is worth $ 0.25 . It follows that y quarters are worth 0.25 y dollars. It is given that these 18 coins are worth $ 3 . It follow that x dimes and y quarters are worth 3 dollars. Represent this mathematically: 0.1 x + 0.25 y = 3 . Multiply the equation x + y = 18 by 0.1 : 0.1 x + 0.1 y = 1.8 Subtract the equation obtained above from the equation 0.1 x + 0.25 y = 3 and solve the resulting equation for y . 0.1 x + 0.25 y − 0.1 x + 0.1 y = 3 − 1.8 0.1 x + 0.25 y − 0.1 x − 0.1 y = 1.2 0.15 y = 1.2 y = 1.2 0.15 y = 8 Substitute y = 8 in the equation x + y = 18 and solve the resulting equation for x : x + y = 18 x + 8 = 18 x = 18 − 8 x = 10 Conclusion: Georgia has x = 10 dimes and y = 8 quarters. Thus, she has 2 dimes more than quarters.

How many more dimes than quarters does Georgia have in the given situation. Georgia has 2 dimes more than quarters. Given: Georgia has only dimes and quarters. She has total 18 coins that are worth $ 3 . Concepts Used: Modelling a situational problem mathematically. Solving a pair of linear equations in two variables using elimination method. Calculations: Let number of dimes be x and the number of quarters be y . The total number of coins is given to be 18 . Represent this mathematically: x + y = 18 . Recall that 1 dime is worth $ 0.10 . It follows that x dimes are worth 0.1 x dollars. Also recall that 1 quarter is worth $ 0.25 . It follows that y quarters are worth 0.25 y dollars. It is given that these 18 coins are worth $ 3 . It follow that x dimes and y quarters are worth 3 dollars. Represent this mathematically: 0.1 x + 0.25 y = 3 . Multiply the equation x + y = 18 by 0.1 : 0.1 x + 0.1 y = 1.8 Subtract the equation obtained above from the equation 0.1 x + 0.25 y = 3 and solve the resulting equation for y . 0.1 x + 0.25 y − 0.1 x + 0.1 y = 3 − 1.8 0.1 x + 0.25 y − 0.1 x − 0.1 y = 1.2 0.15 y = 1.2 y = 1.2 0.15 y = 8 Substitute y = 8 in the equation x + y = 18 and solve the resulting equation for x : x + y = 18 x + 8 = 18 x = 18 − 8 x = 10 Conclusion: Georgia has x = 10 dimes and y = 8 quarters. Thus, she has 2 dimes more than quarters.

Solution Summary: The author explains how to determine how many more dimes than quarters Georgia has in the given situation.

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

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Expert Solution & Answer

Chapter 3.2, Problem 70PPSE

Solution

To Determine:

How many more dimes than quarters does Georgia have in the given situation.

Georgia has 2 dimes more than quarters.

Given:

Georgia has only dimes and quarters.

She has total 18 coins that are worth $3 .

Concepts Used:

Modelling a situational problem mathematically.

Solving a pair of linear equations in two variables using elimination method.

Calculations:

Let number of dimes be x and the number of quarters be y .

The total number of coins is given to be 18 . Represent this mathematically: x+y=18 .

Recall that 1 dime is worth $0.10 . It follows that x dimes are worth 0.1x dollars. Also recall that 1 quarter is worth $0.25 . It follows that y quarters are worth 0.25y dollars. It is given that these 18 coins are worth $3 . It follow that x dimes and y quarters are worth 3 dollars. Represent this mathematically: 0.1x+0.25y=3 .

Multiply the equation x+y=18 by 0.1 :

0.1x+0.1y=1.8

Subtract the equation obtained above from the equation 0.1x+0.25y=3 and solve the resulting equation for y .

0.1x+0.25y−0.1x+0.1y=3−1.80.1x+0.25y−0.1x−0.1y=1.20.15y=1.2y=1.20.15y=8

Substitute y=8 in the equation x+y=18 and solve the resulting equation for x :

x+y=18x+8=18x=18−8x=10

Conclusion:

Georgia has x=10 dimes and y=8 quarters. Thus, she has 2 dimes more than quarters.

Chapter 3.2, Problem 69PPSE

Chapter 3.2, Problem 71PPSE

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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