CHAPTER 7 Systems of Equations Slide 2Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 7.1Systems of Equations in Two Variables 7.2The Substitution Method 7.3The Elimination Method 7.4Applications and Problem Solving 7.5Applications with Motion

OBJECTIVES 7.4 Applications and Problem Solving Slide 3Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. aSolve applied problems by translating to a system of two equations in two variables.

EXAMPLE Maya sells concessions at a local sporting event. In one hour, she sells 72 drinks. The drink sizes are small, which sells for $2 each, and large, which sells for $3 each. If her total sales revenue was $190, how many of each size did she sell? 7.4 Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. AApplications of Systems of Equations (continued) Slide 4Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 1. Familiarize. Suppose that of the 72 drinks, 20 where small and 52 were large. The 72 drinks would then amount to a total of 20($2) + 52($3) = $196. Although our guess is incorrect, checking the guess has familiarized us with the problem. We let s = the number of small drinks and l = the number of large drinks. 7.4 Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. AApplications of Systems of Equations (continued) Slide 5Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 2. Translate. Since a total of 72 drinks were sold, we must have s + l = 72. To find a second equation, we reword some information and focus on the drinks sold. Rewording and Translating: Income from income from small drinks plus large drinks totaled 190 2s + 3l = Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. AApplications of Systems of Equations Slide 6Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE We have the following system: 7.4 Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. AApplications of Systems of Equations (continued) Slide 7Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 3. Solve. Next, we replace l in equation (2) with l = 72  s: 7.4 Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. AApplications of Systems of Equations (continued) Slide 8Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE We find l by substituting 26 for s in equation (3): l = 72  s = 72  26 = Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. AApplications of Systems of Equations Slide 9Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 4. Check. If Maya sold 26 small and 46 large drinks, she would have sold 72 drinks, for a total of 26($2) + 46($3) = $52 + $138 = $ State. Maya sold 26 small drinks and 46 large drinks. 7.4 Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. AApplications of Systems of Equations Slide 10Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE A cookware consultant sells two sizes of pizza stones. The circular stone sells for $26 and the rectangular one sells for $34. In one month she sold 37 stones. If she made a total of $1138 from the sale of the pizza stones, how many of each size did she sell? 7.4 Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. BApplications and Problem Solving (continued) Slide 11Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 1. Familiarize. When faced with a new problem, it is often useful to compare it to a similar problem that you have already solved. Here instead of $2 and $3 drinks, we are counting pizza stones. We let c = the circular stone and r = the rectangular stone. 2. Translate. Since a total of 37 stones were sold, we havec + r = Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. BApplications and Problem Solving (continued) Slide 12Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE c + r = 37 26c + 34r = 1138 Presenting the information in a table can be helpful. $113834r26cMoney Paid 37rc Number of pans $34$26Cost per pan TotalRectangularCircular 7.4 Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. BApplications and Problem Solving (continued) Slide 13Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE We have translated to a system of equations: c + r = 37 26c + 34r = Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. BApplications and Problem Solving (continued) Slide 14Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 3. Carry out. The system can be solved using elimination:  26c  26r =  c + 34r = r = 176 r = 22 Multiply equation (1) by  26 Equation (2) Adding Dividing both sides by Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. BApplications and Problem Solving (continued) Slide 15Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE To solve for c, we substitute 22 for r: c + r = 37 c + 22 = 37 c = Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. BApplications and Problem Solving (continued) Slide 16Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 4. Check. If r = 22 and c = 15, a total of 37 stones were sold. The amount paid was 22($34) + 15($26) = $1138. The numbers check. 5. State. The consultant sold 22 rectangular pizza stones and 15 circular pizza stones. 7.4 Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. BApplications and Problem Solving Slide 17Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 7.4 Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. CApplications and Problem Solving (continued) Slide 18Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. A coffee shop is considering a new mixture of coffee beans. It will be created with Italian Roast beans costing $9.95 per pound and the Venezuelan Blend beans costing $11.25 per pound. The types will be mixed to form a 60-lb batch that sells for $10.50 per pound. How many pounds of each type of bean should go into the blend?

EXAMPLE 7.4 Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. CApplications and Problem Solving (continued) Slide 19Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 1. Familiarize. This problem is similar to our previous examples. Instead of pizza stones we have coffee blends. We have two different prices per pound. Instead of knowing the total amount paid, we know the weight and price per pound of the new blend being made.

EXAMPLE 7.4 Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. CApplications and Problem Solving (continued) Slide 20Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. Let i represent the number of pounds of Italian roast and v represent the number of pounds of Venezuelan blend. 2. Translate. Since a 60-lb batch is being made, we have i + v = 60. Present the information in a table.

EXAMPLE i + v = i v = Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. CApplications and Problem Solving (continued) Slide 21Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. ItalianVenezuelanNew Blend Number of pounds iv60 Price per pound $9.95$11.25$10.50 Value of beans 9.95i11.25v630

EXAMPLE We have translated to a system of equations: i + v = 60 (1) 9.95i v = 630 (2) 7.4 Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. CApplications and Problem Solving (continued) Slide 22Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 3. Carry out. When equation (1) is solved for v, we have v = 60  i. We then substitute for v in equation (2). 9.95i (60  i) = i  11.25i = 630  1.30i =  45 i = 34.6 If i = 34.6, we see from equation (1) that v = Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. CApplications and Problem Solving (continued) Slide 23Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

EXAMPLE 7.4 Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. CApplications and Problem Solving (continued) Slide 24Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 4. Check. If 34.6 lb of Italian Roast and 25.4 lb of Venezuelan Blend are mixed, a 60-lb blend will result. The value of 34.6 lb of Italian beans is 34.6($9.95), or $ The value of 25.4 lb of Venezuelan Blend is 25.4($11.25), or $285.75, so the value of the blend is $ A 60-lb blend priced at $10.50 a pound is also worth $630, so our answer checks.

EXAMPLE 7.4 Applications and Problem Solving a Solve applied problems by translating to a system of two equations in two variables. CApplications and Problem Solving Slide 25Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. 5. State. The new blend should be made from 34.6 pounds of Italian Roast beans and 25.4 pounds of Venezuelan Blend beans.

7.4 Applications and Problem Solving Problem-Solving Tip Slide 26Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc. When solving problems, see if they are patterned or modeled after other problems that you have studied.