Systems of Linear Equations

Transcription

1 4 Systems of Linear Equations Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide 1 1-1

2 4.3 Applications of R.1 Systems Fractions of Linear Equations Objectives 1. Solve geometric problems using two variables. 2. Solve money problems using two variables. 3. Solve mixture problems using two variables. 4. Solve distance-rate problems using two variables. 5. Solve problems with three variables using a system of three equations. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide 2 1-2

3 Solving an Applied Problem by Writing a System of Equations Step 1 Read the problem, several times if necessary. What information is given? What is to be found? This is often stated in the last sentence. Step 2 Assign variables to represent the unknown values. Use a sketch, diagram, or table, as needed. Step 3 Write a system of equations using the variable expressions. Step 4 Solve the system of equations. Step 5 State the answer to the problem. Label it appropriately. Does it seem reasonable? Step 6 Check the answer in the words of the original problem. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide 3 1-3

4 Example 1 Finding 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 Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide 4 1-4

5 Finding 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: 2L + 2W = 800. Because the length is 20 ft more than three times its width, we have The system is, therefore, L = 3W L + 2W = 800 (1) L = 3W (2) Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide 5 1-5

6 EXAMPLE Continued. 1 Finding 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 3W + 20 for L, in equation (1), and solve for W. 2L + 2W = 800 (1) 2(3W + 20) + 2W = 800 Let L = 3W W W = 800 8W + 40 = 800 Distributive property Combine terms. 8W = 760 Subtract 40. W = 95 Divide by 8. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide 6 1-6

7 Finding 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 2L + 2W = 800 and found W = 95. Now, let W = 95 in the equation L = 3W + 20 to find L. L = 3W + 20 L = 3(95) + 20 Let W = 95. L = 305 Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide 7 1-7

8 EXAMPLE Continued. 1 Finding 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. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide 8 1-8

9 Example 2 Solving a Problem about Prices At a local restaurant, the price of 1 soft drink and 2 hamburgers is $12.15 and EXAMPLE 2 the price of 5 soft drinks and 4 hamburgers is $ 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. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide 9 1-9

10 Solving a Problem about Prices At a local restaurant, the price of 1 soft drink and 2 hamburgers is $12.15 and EXAMPLE 2 the price of 5 soft drinks and 4 hamburgers is $ 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 s + 2h = By similar reasoning, the second equation is Therefore, the system is 5s + 4h = s + 2h = (1) 5s + 4h = (2) Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

11 Solving a Problem about Prices At a local restaurant, the price of 1 soft drink and 2 hamburgers is $12.15 and EXAMPLE 2 the price of 5 soft drinks and 4 hamburgers is $ Find the price of a single hamburger and a soft drink. Step 4 Solve the system of equations. s + 2h = (1) 5s + 4h = (2) 5s 10h = Multiply each side of (1) by 5. 5s + 4h = (2) 6h = Add. h = 4.75 Divide by 6. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

12 Solving a Problem about Prices At a local restaurant, the price of 1 soft drink and 2 hamburgers is $12.15 and EXAMPLE 2 the price of 5 soft drinks and 4 hamburgers is $ Find the price of a single hamburger and a soft drink. Step 4 Solve the system of equations. We just found h = Now, let h = 4.75 in the equation s + 2h = to find s. s + 2h = (1) s + 2(4.75) = Let h = s = Multiply. s = 2.65 Subtract Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

13 Solving a Problem about Prices At a local restaurant, the price of 1 soft drink and 2 hamburgers is $12.15 and EXAMPLE 2 the price of 5 soft drinks and 4 hamburgers is $ 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. Step 6 Check that these values satisfy the conditions stated in the problem. 5(2.65) 1(2.65) + 4(4.75) 2(4.75) = $32.25 $12.15 The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

14 Example 3 Solving 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 1 Read the problem. Two solutions of different strengths are being mixed together to get a specific amount of a solution with an inbetween strength. Step 2 Assign a variable. Let x = the number of ounces of 10% solution and y = the number of ounces of 25% solution. 25% + = 10% 25% 22% 10% x oz y oz 40 oz Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

15 Solving 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 2 Assign a variable. Let x = the number of ounces of 10% solution and y = the number of ounces of 25% solution. Percent (as a decimal) 10% = % = % = 0.22 Number of Ounces x y 40 Ounces of Pure Acid 0.10x 0.25y 0.22(40) Step 3 Write a system of equations. + x + y = = 40 10% 25% 22%.10x x oz y oz +.25y = oz (1) (2) Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

16 Solving 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 4 Solve the system. x y 40 + = (1) 0.10x 0.25y = (2) 10x 10y = 10x 25y = Multiply each side of (1) by 10. Multiply each side of (2) by y = 480 Add. y = 32 Divide by 15. Because y = 32 and x + y = 40, x = 8. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

17 Solving 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 5 State the answer. The desired mixture will require 8 oz of the 10% solution and 32 oz of the 25% solution. Step 6 Check that these values satisfy both equations of the system. Percent (as a decimal) Number of Ounces 8 32 Ounces of Pure Acid Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

18 Example 4 Solving a Motion Problem 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 1 Read the problem again. Given the distances traveled, we need to find the speed of each vehicle. Step 2 Assign variables. Let x = the speed of the car Since d = rt, and y = the speed of the bus. t = d r. distance (d) rate (r) time (t) Car 310 x 310 x Bus 290 y 290 y Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

19 Solving a Motion Problem 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 3 Write 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, x = y + 4. Both vehicles travel for the same time, so from the table x = y. Multiplying both sides by xy gives 310y = 290x. distance (d) rate (r) time (t) Car 310 x 310 x Bus 290 y 290 y Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

20 Solving a Motion Problem 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 4 Solve the system of equations using substitution. x = y + 4 (1) 310y = 290x (2) 310y = 290x (2) 310y = 290(y + 4) Let x = y y = 290y Distributive property 20y = 1160 Subtract 290y. y = 58 Divide by 20. Because x = y + 4, the value of x is = 62. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

21 Solving a Motion Problem 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 5 State the answer. The car s speed is 62 mph, and the speed of the bus is 58 mph. Step 6 Check. This is especially important since one of the equations had variable denominators. Car: t = d r = Bus: t = d r = = 5 = 5 Times are equal. Since = 4, the conditions of the problem are satisfied. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

22 Example 6 Solving 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 $ How many loaves of each type of bread were sold? Step 1 Read the problem again. There are three unknowns. Step 2 Assign 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. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

23 Example 6 Solving 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 $ How many loaves of each type of bread were sold? Step 3 Write a system of equations. x = loaves of honey wheat, y = loaves of sunflower, z = loaves of French bread y = 2x, or y 2x = 0 (1) z = y 3, or z y = 2 (2) 2.69x y z = (3) Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

24 EXAMPLE Continued. 5 Solving 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 $ How many loaves of each type of bread were sold? Step 4 Solve the system of three equations using substitution. y = 2x (1) z = y 3 (2) z = 2x 3 Let y = 2x. (4) Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

25 Solving 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 $ How many loaves of each type of bread were sold? Step 4 Solve the system of three equations using substitution. y = 2x z = 2x 3 (1) (4) 269x + 289y + 339z = 9658 Multiply each side (3) by x + 289(2x) + 339(2x 3) = 9658 Substitute in (3). Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

26 EXAMPLE Continued. 5 Solving 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 $ How many loaves of each type of bread were sold? Step 4 Solve the system of three equations using substitution. 269x + 578x + 678x 1017 = x 1017 = 9658 Multiply & distribute. Combine terms. 1525x = 10,675 Add x = 7 Divide by Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

27 Solving 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 $ How many loaves of each type of bread were sold? Step 4 Solve the system of three equations using substitution. Because x = 7 and y = 2x, y = 14. Also, because y = 14 and z = y 3, z = 11. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

28 Solving 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 $ How many loaves of each type of bread were sold? Step 5 State 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. Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide

29 EXAMPLE 5 Solving 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 $ How many loaves of each type of bread were sold? Step 6 Check. 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) (14) (11) = Copyright 2014, 2010, 2006 Pearson Education, Inc. Section 4.3, Slide