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
Expert Solution & Answer
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Chapter 3.2, Problem 9LC
Solution

To explain how to write and solve a systems of equations to find the number of large cups of coffee they bought.

The variables x and y are used to represent the number of regular cups of coffee and large cups of coffee, respectively.

Developing a mathematical model resulted in the following system of equations:

  x+y=5x+1.5y=6 .

Solving the system of equations, it is found that:

  y=2 .

That is, the number of large cups of coffee is found to be 2 .

Given:

Given that the café sells a regular cup of coffee for $1 and a large cup of coffee for $1.50 .

Again, it is given that, Melissa and her friends bought 5 cups of coffee for $6 .

Concept Used:

Method of elimination in the case of linear equations in two variables:

Given a system of linear equations in two variables.

The method of elimination employs the elimination of one variable from the system of equations. Thus, the first step is to determine the variable that is to be eliminated. Then, the equations are multiplied with different non-zero constants and added together in a way so that the targeted variable is eliminated. In some cases, a simple addition would eliminate the variable. But, in general scalar multiplication is needed as said earlier.

Calculation:

Develop a mathematical model of the problem as a system of equations:

Suppose that x and y denote the numbers of regular and large cup of coffee they bought, respectively.

Since they bought a total of 5 cups of coffee, it follows that:

  x+y=5 .

Now, since x is the number of regular cups of coffee and each such cup costs $1 , the total cost of regular cup of coffee in dollars:

  1x .

Similarly, the total cost of large cup of coffee in dollars:

  1.5y

Thus, the total cost of coffee:

  1x+1.5y .

It is given that the total cost on coffee is $6 .

Then, it follows that:

  1x+1.5y=6 .

That is:

  x+1.5y=6 .

Then, a system of linear equations is found:

  x+y=5x+1.5y=6 .

Solve the system of equations using the method of elimination:

Since it is asked to find the number of large cups, the method is applied to eliminate the variable x .

Multiply the second equation x+1.5y=6 with 1 :

  1x+1.5y=16x1.5y=6(1) .

Add equation 1 with the first equation x+y=5 :

  x1.5y=6x+y=50x0.5y=1+ .

Thus, it is found that:

  0x0.5y=1 .

Find the value of y from 0x0.5y=1 :

  0x0.5y=100.5y=10.5y=1y=10.5y=2 .

Thus, it is found that:

  y=2 .

Thus, the number of large cups of coffee:

  2 .

Conclusion:

The variables x and y are used to represent the number of regular cups of coffee and large cups of coffee, respectively.

Developing a mathematical model resulted in the following system of equations:

  x+y=5x+1.5y=6 .

Solving the system of equations, it is found that:

  y=2 .

That is, the number of large cups of coffee is found to be 2 .

Chapter 3 Solutions

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

Ch. 3.1 - Prob. 11PPSECh. 3.1 - Prob. 12PPSECh. 3.1 - Prob. 13PPSECh. 3.1 - Prob. 14PPSECh. 3.1 - Prob. 15PPSECh. 3.1 - Prob. 16PPSECh. 3.1 - Prob. 17PPSECh. 3.1 - Prob. 18PPSECh. 3.1 - Prob. 19PPSECh. 3.1 - Prob. 20PPSECh. 3.1 - Prob. 21PPSECh. 3.1 - Prob. 22PPSECh. 3.1 - Prob. 23PPSECh. 3.1 - Prob. 24PPSECh. 3.1 - Prob. 25PPSECh. 3.1 - Prob. 26PPSECh. 3.1 - Prob. 27PPSECh. 3.1 - Prob. 28PPSECh. 3.1 - Prob. 29PPSECh. 3.1 - Prob. 30PPSECh. 3.1 - Prob. 31PPSECh. 3.1 - Prob. 32PPSECh. 3.1 - Prob. 33PPSECh. 3.1 - Prob. 34PPSECh. 3.1 - Prob. 35PPSECh. 3.1 - Prob. 36PPSECh. 3.1 - Prob. 37PPSECh. 3.1 - Prob. 38PPSECh. 3.1 - Prob. 39PPSECh. 3.1 - Prob. 40PPSECh. 3.1 - Prob. 41PPSECh. 3.1 - Prob. 42PPSECh. 3.1 - Prob. 43PPSECh. 3.1 - Prob. 44PPSECh. 3.1 - Prob. 45PPSECh. 3.1 - Prob. 46PPSECh. 3.1 - Prob. 47PPSECh. 3.1 - Prob. 48PPSECh. 3.1 - Prob. 49PPSECh. 3.1 - Prob. 50PPSECh. 3.1 - Prob. 51PPSECh. 3.1 - Prob. 52PPSECh. 3.1 - Prob. 53PPSECh. 3.1 - Prob. 54PPSECh. 3.1 - Prob. 55PPSECh. 3.1 - 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Prob. 33PPSECh. 3.6 - Prob. 34PPSECh. 3.6 - Prob. 35PPSECh. 3.6 - Prob. 36PPSECh. 3.6 - Prob. 37PPSECh. 3.6 - Prob. 38PPSECh. 3.6 - Prob. 39PPSECh. 3.6 - Prob. 40PPSECh. 3.6 - Prob. 41PPSECh. 3.6 - Prob. 42PPSECh. 3.6 - Prob. 43PPSECh. 3.6 - Prob. 44PPSECh. 3.6 - Prob. 45PPSECh. 3.6 - Prob. 46PPSECh. 3.6 - Prob. 47PPSECh. 3.6 - Prob. 48PPSECh. 3.6 - Prob. 49PPSECh. 3.6 - Prob. 50PPSECh. 3.6 - Prob. 51PPSECh. 3.6 - Prob. 52PPSECh. 3.6 - Prob. 53PPSECh. 3.6 - Prob. 54PPSECh. 3.6 - Prob. 55PPSECh. 3.6 - Prob. 56PPSECh. 3.6 - Prob. 57PPSECh. 3.6 - Prob. 58PPSECh. 3 - Prob. 1GRCh. 3 - Prob. 2GRCh. 3 - Prob. 3GRCh. 3 - Prob. 4GRCh. 3 - Prob. 5GRCh. 3 - Prob. 6GRCh. 3 - Prob. 7GRCh. 3 - Prob. 8GRCh. 3 - Prob. 9GRCh. 3 - Prob. 10GRCh. 3 - Prob. 11GRCh. 3 - Prob. 12GRCh. 3 - Prob. 13GRCh. 3 - Prob. 14GRCh. 3 - Prob. 15GRCh. 3 - Prob. 16GRCh. 3 - Prob. 17GRCh. 3 - Prob. 18GRCh. 3 - Prob. 19GRCh. 3 - Prob. 20GRCh. 3 - Prob. 21GRCh. 3 - Prob. 22GRCh. 3 - Prob. 23GRCh. 3 - Prob. 1MCQCh. 3 - Prob. 2MCQCh. 3 - 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