1. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 5.3 - 1

2. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 5.3 - 2 Systems of Linear Equations Chapter 5

3. Copyright © 2010 Pearson Education, Inc. All rights reserved Sec 5.3 - 3 5.3 Applications of Systems of Linear Equations

4. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 4 5.3 Applications of Systems of Linear Equations Objectives 1.Solve geometry problems using two variables. 2.Solve money problems using two variables. 3.Solve mixture problems using two variables. 4.Solve distance-rate-time problems using two variables. 5.Solve problems with three variables using a system of three equations.

5. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 5 5.3 Applications of Systems of Linear Equations Solving an Applied Problem by Writing a System of Equations Step 1 Read the problem, several times if necessary, until you understand what is given and what is to be found. Step 2 Assign variables to represent the unknown values, using diagrams or tables as needed. Write down what each variable represents. Step 3 Write a system of equations that relates the unknowns. Step 4 Solve the system of equations. Step 5 State the answer to the problem. Does it seem reasonable? Step 6 Check the answer in the words of the original problem.

6. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 6 5.3 Applications of Systems of Linear Equations EXAMPLE 1Finding the Dimensions of a Parking Lot The length of a rectangular parking lot is 20 ft more than three times its width. The perimeter of the parking lot is 800 ft. What are the dimensions of the parking lot? Step 1 Read the problem again. We must find the dimensions of the parking lot. Step 2 Assign variables. Let L = the length and W = the width. L W

7. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 7 5.3 Applications of Systems of Linear Equations EXAMPLE 1Finding the Dimensions of a Parking Lot The length of a rectangular parking lot is 20 ft more than three times its width. The perimeter of the parking lot is 800 ft. What are the dimensions of the parking lot? Step 3 Write a system of equations. Because the perimeter is 800 ft, we find one equation by using the perimeter formula: 2 L + 2 W = 800. Because the length is 20 ft more than three times its width, we have L = 3 W + 20. The system is, therefore, 2 L + 2 W = 800(1) L = 3 W + 20.(2)

8. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 8 5.3 Applications of Systems of Linear Equations EXAMPLE 1Finding the Dimensions of a Parking Lot The length of a rectangular parking lot is 20 ft more than three times its width. The perimeter of the parking lot is 800 ft. What are the dimensions of the parking lot? Step 4 Solve the system of equations. We substitute 3 W + 20 for L, in equation (1), and solve for W. 2 L + 2 W = 800 (1) 2(3 W + 20) + 2 W = 800 Let L = 3 W + 20. 6 W + 40 + 2 W = 800 Distributive property 8 W + 40 = 800 Combine terms. 8 W = 760 Subtract 40. W = 95Divide by 8.

9. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 9 5.3 Applications of Systems of Linear Equations EXAMPLE 1Finding the Dimensions of a Parking Lot The length of a rectangular parking lot is 20 ft more than three times its width. The perimeter of the parking lot is 800 ft. What are the dimensions of the parking lot? Step 4 Solve the system of equations. We just solved the equation 2 L + 2 W = 800 and found W = 95. Now, let W = 95 in the equation L = 3 W + 20 to find L. L = 3 W + 20 L = 3(95) + 20 Let W = 95. L = 305

10. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 10 5.3 Applications of Systems of Linear Equations EXAMPLE 1Finding the Dimensions of a Parking Lot The length of a rectangular parking lot is 20 ft more than three times its width. The perimeter of the parking lot is 800 ft. What are the dimensions of the parking lot? Step 5 State the answer. The length is 305 ft and the width is 95 ft. Step 6 Check. The perimeter is 2(305) + 2(95) = 800 ft, and the length, 305 ft, is 20 ft more than three times the width, since 3(95) + 20 = 305. The answer is correct.

11. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 11 EXAMPLE 2 Solving a Problem about Prices At a local restaurant, the price of 1 soft drink and 2 hamburgers is $12.15 and the price of 5 soft drinks and 4 hamburgers is $32.25. Find the price of a single hamburger and a soft drink. Step 1 Read the problem again. There are two unknowns. Step 2 Assign variables. Let s represent the price of one soft drink and h represent the price of one hamburger. EXAMPLE 2 5.3 Applications of Systems of Linear Equations

12. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 12 EXAMPLE 2 Solving a Problem about Prices At a local restaurant, the price of 1 soft drink and 2 hamburgers is $12.15 and the price of 5 soft drinks and 4 hamburgers is $32.25. Find the price of a single hamburger and a soft drink. Step 3 Write a system of equations. Because one soft drink and 2 hamburgers cost a total of $12.15, one equation for the system is EXAMPLE 2 5.3 Applications of Systems of Linear Equations s + 2 h = 12.15. By similar reasoning, the second equation is 5 s + 4 h = 32.25. Therefore, the system is s + 2 h = 12.15 (1) 5 s + 4 h = 32.25.(2)

13. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 13 EXAMPLE 2 Solving a Problem about Prices At a local restaurant, the price of 1 soft drink and 2 hamburgers is $12.15 and the price of 5 soft drinks and 4 hamburgers is $32.25. Find the price of a single hamburger and a soft drink. Step 4 Solve the system of equations. EXAMPLE 2 5.3 Applications of Systems of Linear Equations –5 s – 10 h = –60.75 Multiply each side of (1) by –5. 5 s + 4 h = 32.25(2) –6 h = –28.50 Add. h = 4.75 Divide by –6. s + 2 h = 12.15 (1) 5 s + 4 h = 32.25(2)

14. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 14 EXAMPLE 2 Solving a Problem about Prices At a local restaurant, the price of 1 soft drink and 2 hamburgers is $12.15 and the price of 5 soft drinks and 4 hamburgers is $32.25. Find the price of a single hamburger and a soft drink. Step 4 Solve the system of equations. We just found h = 4.75. Now, let h = 4.75 in the equation s + 2 h = 12.15 to find s. EXAMPLE 2 5.3 Applications of Systems of Linear Equations s + 2 h = 12.15 (1) s + 2(4.75) = 12.15 Let h = 4.75. s + 9.50 = 12.15 Multiply. s = 2.65 Subtract 9.50.

15. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 15 5.3 Applications of Systems of Linear Equations EXAMPLE 2 Solving a Problem about Prices At a local restaurant, the price of 1 soft drink and 2 hamburgers is $12.15 and the price of 5 soft drinks and 4 hamburgers is $32.25. Find the price of a single hamburger and a soft drink. Step 5 State the answer. The price of a single soft drink is $2.65 and the price of a hamburger is $4.75. EXAMPLE 2 Step 6 Check that these values satisfy the conditions stated in the problem. 5(2.65) 4(4.75) = $32.25+ 1(2.65) 2(4.75) = $12.15+

16. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 16 5.3 Applications of Systems of Linear Equations EXAMPLE 3Solving a Mixture Problem Step 1Read the problem. Two solutions of different strengths are being mixed together to get a specific amount of a solution with an “in- between” strength. How many ounces each of 10% hydrochloric acid and 25% hydrochloric acid must be combined to get 40 oz of solution that is 22% hydrochloric acid? Step 2Assign a variable. Let x = the number of ounces of 10% solution and y = the number of ounces of 25% solution. + = 10%25% 10% 25% x oz y oz 40 oz 22%

17. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 17 Step 2Assign a variable. Let x = the number of ounces of 10% solution and y = the number of ounces of 25% solution. 5.3 Applications of Systems of Linear Equations EXAMPLE 3Solving a Mixture Problem How many ounces each of 10% hydrochloric acid and 25% hydrochloric acid must be combined to get 40 oz of solution that is 22% hydrochloric acid? Percent (as a decimal)Ounces of Pure Acid 0.10x Number of Ounces 0.25y 0.22(40) x y 40 Step 3Write a system of equations. += + = 10%25% 22% x oz y oz 40 oz.10 x.25 y 8.8 += (1) (2) x y 40 10% = 0.10 25% = 0.25 22% = 0.22

18. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 18 Step 4Solve the system. 5.3 Applications of Systems of Linear Equations EXAMPLE 3Solving a Mixture Problem How many ounces each of 10% hydrochloric acid and 25% hydrochloric acid must be combined to get 40 oz of solution that is 22% hydrochloric acid? x y 40 += 0.10 x 0.25 y 8.8 += (1) (2) –10 x –10 y –400 += 10 x 25 y 880 += Multiply each side of (1) by –10. Multiply each side of (2) by 100. 15 y 480 = Add. y 32 = Divide by 15. Because y = 32 and x + y = 40, x = 8.

19. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 19 Step 5State the answer. The desired mixture will require 8 oz of the 10% solution and 32 oz of the 25% solution. 5.3 Applications of Systems of Linear Equations EXAMPLE 3Solving a Mixture Problem How many ounces each of 10% hydrochloric acid and 25% hydrochloric acid must be combined to get 40 oz of solution that is 22% hydrochloric acid? Step 6Check that these values satisfy both equations of the system. Percent (as a decimal)Ounces of Pure AcidNumber of Ounces 0.10 0.25 8 32 0.22 0.8 8 408.8

20. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 20 5.3 Applications of Systems of Linear Equations EXAMPLE 4Solving a Motion Problem Step 1Read the problem again. Given the distances traveled, we need to find the speed of each vehicle. A car travels at 310 miles in the same time that a bus travels 290 miles. If the speed of the car is 4 mph faster than the speed of the bus, find both speeds. Step 2Assign variables. Let x = the speed of the car and y = the speed of the bus. distance ( d )time ( t )rate ( r ) Car Bus x y 310 290 Since d = rt, t =. d r 310 x 290 y

21. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 21 5.3 Applications of Systems of Linear Equations EXAMPLE 4Solving a Motion Problem Step 3Write a system of equations. The problem states that the car travels 4 mph faster than the bus. Since the two speeds are x and y, A car travels at 310 miles in the same time that a bus travels 290 miles. If the speed of the car is 4 mph faster than the speed of the bus, find both speeds. x = y + 4. Both vehicles travel for the same time, so from the table 310 x 290 y =.=. distance ( d )time ( t )rate ( r ) Car Bus x y 310 290 310 x 290 y Multiplying both sides by xy gives 310 y = 290 x.

22. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 22 5.3 Applications of Systems of Linear Equations EXAMPLE 4Solving a Motion Problem Step 4Solve the system of equations using substitution. A car travels at 310 miles in the same time that a bus travels 290 miles. If the speed of the car is 4 mph faster than the speed of the bus, find both speeds. x = y + 4 (1) 310 y = 290 x (2) 310 y = 290( y + 4)Let x = y + 4. 310 y = 290 y + 1160Distributive property 20 y = 1160 Subtract 290 y. y = 58 Divide by 20. Because x = y + 4, the value of x is 58 + 4 = 62.

23. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 23 5.3 Applications of Systems of Linear Equations EXAMPLE 4Solving a Motion Problem Step 5State the answer. The car’s speed is 62 mph, and the speed of the bus is 58 mph. A car travels at 310 miles in the same time that a bus travels 290 miles. If the speed of the car is 4 mph faster than the speed of the bus, find both speeds. Step 6Check. This is especially important since one of the equations had variable denominators. Car: t = d r = 310 62 = 5 Bus: t = d r = 290 58 = 5 Since 62 – 58 = 4, the conditions of the problem are satisfied. Times are equal.

24. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 24 5.3 Applications of Systems of Linear Equations EXAMPLE 5Solving a Problem Involving Prices Step 1Read the problem again. There are three unknowns. At a local bakery, a loaf of honey wheat bread costs $2.69, a loaf of sunflower bread costs $2.89, and a loaf of French bread costs $3.39. On a recent day, twice as many loaves of sunflower were sold as honey wheat. The number of loaves of French bread sold was three less than the number of loaves of sunflower. Total receipts for these breads were $96.58. How many loaves of each type of bread were sold? Step 2Assign variables to represent the three unknowns. Let x = number of loaves of honey wheat, y = number of loaves of sunflower, and z = number of loaves of French bread.

25. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 25 5.3 Applications of Systems of Linear Equations EXAMPLE 5Solving a Problem Involving Prices Step 3Write a system of equations. At a local bakery, a loaf of honey wheat bread costs $2.69, a loaf of sunflower bread costs $2.89, and a loaf of French bread costs $3.39. On a recent day, twice as many loaves of sunflower were sold as honey wheat. The number of loaves of French bread sold was three less than the number of loaves of sunflower. Total receipts for these breads were $96.58. How many loaves of each type of bread were sold? y = 2 x, or y – 2 x = 0(1) x = loaves of honey wheat, y = loaves of sunflower, z = loaves of French bread z = y – 3, or z – y = –2(2) 2.69 x + 2.89 y + 3.39 z = 96.58(3)

26. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 26 5.3 Applications of Systems of Linear Equations EXAMPLE 5Solving a Problem Involving Prices Step 4Solve the system of three equations using substitution. At a local bakery, a loaf of honey wheat bread costs $2.69, a loaf of sunflower bread costs $2.89, and a loaf of French bread costs $3.39. On a recent day, twice as many loaves of sunflower were sold as honey wheat. The number of loaves of French bread sold was three less than the number of loaves of sunflower. Total receipts for these breads were $96.58. How many loaves of each type of bread were sold? y = 2 x (1) z = y – 3 (2) z = 2 x – 3 Let y = 2 x. (4)

27. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 27 5.3 Applications of Systems of Linear Equations EXAMPLE 5Solving a Problem Involving Prices Step 4Solve the system of three equations using substitution. At a local bakery, a loaf of honey wheat bread costs $2.69, a loaf of sunflower bread costs $2.89, and a loaf of French bread costs $3.39. On a recent day, twice as many loaves of sunflower were sold as honey wheat. The number of loaves of French bread sold was three less than the number of loaves of sunflower. Total receipts for these breads were $96.58. How many loaves of each type of bread were sold? y = 2 x (1) z = 2 x – 3 (4) 269 x + 289 y + 339 z = 9658 Multiply each side (3) by 100. 269 x + 289(2 x ) + 339(2 x – 3) = 9658 Substitute in (3).

28. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 28 5.3 Applications of Systems of Linear Equations EXAMPLE 5Solving a Problem Involving Prices Step 4Solve the system of three equations using substitution. At a local bakery, a loaf of honey wheat bread costs $2.69, a loaf of sunflower bread costs $2.89, and a loaf of French bread costs $3.39. On a recent day, twice as many loaves of sunflower were sold as honey wheat. The number of loaves of French bread sold was three less than the number of loaves of sunflower. Total receipts for these breads were $96.58. How many loaves of each type of bread were sold? 269 x + 578 x + 678 x – 1017 = 9658 Multiply & distribute. 1525 x – 1017 = 9658 Combine terms. 1525 x = 10,675 Add 1017. x = 7 Divide by 1525.

29. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 29 5.3 Applications of Systems of Linear Equations EXAMPLE 5Solving a Problem Involving Prices Step 4Solve the system of three equations using substitution. At a local bakery, a loaf of honey wheat bread costs $2.69, a loaf of sunflower bread costs $2.89, and a loaf of French bread costs $3.39. On a recent day, twice as many loaves of sunflower were sold as honey wheat. The number of loaves of French bread sold was three less than the number of loaves of sunflower. Total receipts for these breads were $96.58. How many loaves of each type of bread were sold? Because x = 7 and y = 2 x, y = 14. Also, because y = 14 and z = y – 3, z = 11.

30. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 30 5.3 Applications of Systems of Linear Equations EXAMPLE 5Solving a Problem Involving Prices Step 5State the answer. The solution set is { (7, 14, 11) }, meaning that 7 loaves of honey wheat, 14 loaves of sunflower, and 11 loaves of French bread were sold. At a local bakery, a loaf of honey wheat bread costs $2.69, a loaf of sunflower bread costs $2.89, and a loaf of French bread costs $3.39. On a recent day, twice as many loaves of sunflower were sold as honey wheat. The number of loaves of French bread sold was three less than the number of loaves of sunflower. Total receipts for these breads were $96.58. How many loaves of each type of bread were sold?

31. Copyright © 2010 Pearson Education, Inc. All rights reserved. Sec 5.3 - 31 5.3 Applications of Systems of Linear Equations EXAMPLE 5Solving a Problem Involving Prices At a local bakery, a loaf of honey wheat bread costs $2.69, a loaf of sunflower bread costs $2.89, and a loaf of French bread costs $3.39. On a recent day, twice as many loaves of sunflower were sold as honey wheat. The number of loaves of French bread sold was three less than the number of loaves of sunflower. Total receipts for these breads were $96.58. How many loaves of each type of bread were sold? Step 6Check. Since 14 = 2(7), the number of loaves of sunflower sold is twice the number of loaves as honey wheat. Also, 14 – 3 = 11, so the number of loaves of French bread is three less than the number of loaves of sunflower. The total from the receipts is $96.58 as stated. 2.69(7) + 2.89(14) + 3.39(11) = 96.58.