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1 Equations and Inequalities. 1.2 Applications of Linear Equations.

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Presentation on theme: "1 Equations and Inequalities. 1.2 Applications of Linear Equations."— Presentation transcript:

1 1 Equations and Inequalities

2 1.2 Applications of Linear Equations

3 Objectives 1.Solve Number Problems 2.Solve Geometric Problems 3.Solve Investment Problems 4.Solve Break-Point Analysis Problems 5.Solve Shared-Work Problems 6.Solve Mixture Problems 7.Solve Uniform Motion Problems

4 Mathematical Modeling In this section, we will use the equation-solving techniques discussed in the previous section to solve applied problems (often called word problems). To solve these problems, we must translate the verbal description of the problem into an equation. The process of finding the equation that describes the words of the problem is called mathematical modeling. The equation itself is often called a mathematical model of the situation described in the word problem.

5 Strategy for Modeling 1.Analyze the problem to see what you are to find. Often, drawing a diagram or making a table will help you visualize the facts. 2.Pick a variable to represent the quantity that is to be found, and write a sentence telling what that variable represents. Express all other quantities mentioned in the problem as expressions involving this single variable. 3.Find a way to express a quantity in two different ways. This might involve a formula from geometry, finance, or physics.

6 Strategy for Modeling 4.Form an equation indicating that the two quantities found in Step 3 are equal. 5.Solve the equation. 6.Answer the questions asked in the problem. 7.Check the answers in the words of the problem.

7 1. Solve Number Problems

8 Example 1 A student has scores of 74%, 78%, and 70% on three exams. What score is needed on a fourth exam for the student to earn an average grade of 80%?

9 Example 1 – Solution To find an equation that models the problem, we can let x represent the required grade on the fourth exam. The average grade will be one-fourth of the sum of the four grades. We know this average is to be 80.

10 Example 1 – Solution We can solve this equation for x.

11 Example 1 – Answer To earn an average of 80%, the student must score 98% on the fourth exam.

12 2. Solve Geometric Problems

13 Example 2 A city ordinance requires a man to install a fence around the swimming pool shown in the figure. He wants the border around the pool to be of uniform width. If he has 154 feet of fencing, find the width of the border.

14 Example 2 – Solution We can let x represent the width of the border. The distance around the large rectangle, called its perimeter, is given by the formula P = 2l + 2w, where is the l length, 20 + 2x, and w is the width, 16 + 2x. Since the man has 154 feet of fencing, the perimeter will be 154 feet.

15 Example 2 – Solution To find an equation that models the problem, we substitute these values into the formula for perimeter.

16 Example 2 – Answer

17 3. Solve Investment Problems

18 Example 3 A woman invested $10,000, part at 9% and the rest at 14%. If the annual income from these investments is $1,275, how much did she invest at each rate?

19 Example 3 – Solution We can let x represent the amount invested at 9%. Then 10,000 – x represents the amount invested at 14%. Since the annual income from any simple- interest investment is the product of the interest rate and the amount invested, we have the following information.

20 Example 3 – Solution Type of investment RateAmount invested Interest earned 9% investment 0.09x0.09x 14% investment 0.1410,000 – x0.14(10,000 – x)

21 Example 3 – Solution The total income from these two investments can be expressed in two ways: as $1,275 and as the sum of the incomes from the two investments.

22 Example 3 – Solution We can solve this equation for x.

23 Example 3 – Answer The amount invested at 9% was $2,500, and the amount invested at 14% was $7,500. These amounts are correct, because 9% of $2,500 is $225, 14% of $7,500 is $1,050, and the sum of these amounts is $1,275.

24 4. Solve Break-Point Analysis Problems

25 Costs Running a machine involves two costs— setup costs and unit costs. Setup costs include the cost of installing a machine and preparing it to do a job. Unit cost is the cost to manufacture one item, which includes the costs of material and labor.

26 Example 4 Suppose that one machine has a setup cost of $400 and a unit cost of $1.50, and a second machine has a setup cost of $500 and a unit cost of $1.25. Find the break point (the number of units manufactured at which the cost on each machine is the same).

27 Example 4 – Solution We can let x represent the number of items to be manufactured. The cost c 1 of using machine 1 is c 1 = 400 + 1.5x and the cost c 2 of using machine 2 is c 2 = 500 + 1.25x

28 Example 4 – Solution The break point occurs when these two costs are equal.

29 Example 4 – Solution We can solve this equation for x.

30 Example 4 – Answer The break point is 400 units. This result is correct, because it will cost the same amount to manufacture 400 units with either machine. c 1 = $400 + $1.5(400) = $1,000 c 2 = $500 + $1.25(400) = $1,000

31 5. Solve Shared-Work Problems

32 Example 5 The Toll Way Authority needs to pave 100 miles of interstate highway before freezing temperatures come in about 60 days. Sjostrom and Sons has estimated that it can do the job in 110 days. Scandroli and Sons has estimated that it can do the job in 140 days. If the authority hires both contractors, will the job get done in time?

33 Example 5 – Solution Since Sjostrom can do the job in 110 days, they can do 1/110 of the job in one day, and since Scandroli can do the job in 140 days, they can do 1/140 of the job in one day. If we let n represent the number of days it will take to pave the highway if both contractors work together, they can do 1/n of the job in one day.

34 Example 5 – Solution The work that they can do together in one day is the sum of what each can do in one day.

35 Example 5 – Solution We can solve this equation for n.

36 Example 5 – Answer It will take the contractors about 62 days to pave the highway. With any luck, the job will be done in time.

37 6. Solve Mixture Problems

38 Example 6 A container is partially filled with 20 liters of whole milk containing 4% butterfat. How much 1% milk must be added to obtain a mixture that is 2% butterfat?

39 Example 6 – Solution Since the first container contains 20 liters of 4% milk, it contains 0.04(20) liters of butterfat. To this amount, we will add the contents of the second container, which holds 0.01l liters of butterfat. The sum of these two amounts will equal the number of liters of butterfat in the third container, which is 0.02(20 + l) liters of butterfat.

40 Example 6 – Solution We can solve this equation for l.

41 Example 6 – Answer To dilute the 20 liters of 4% milk to a 2% mixture, 40 liters of 1% milk must be added. To check, we note that the final mixture contains 0.02(60) = 1.2 liters of pure butterfat, and that this is equal to the amount of pure butterfat in the 4% milk and the 1% milk; 0.04(20) + 0.01(40) = 1.2 liters.

42 7. Solve Uniform Motion Problems

43 Example 7 A man leaves home driving at the rate of 50 mph. When his daughter discovers that he has forgotten his wallet, she drives after him at the rate of 65 mph. How long will it take her to catch her dad if he had a 15-minute head start?

44 Example 7 – Solution Uniform motion problems are based on the formula d = rt, where d is the distance, r is the rate, and t is the time. Let t represents the number of hours the daughter must drive to overtake her father. Because the father has a 15-minute, or ¼ hour, head start, he has been on the road for (t + ¼) hours.

45 Example 7 – Solution We can set up the following equation and solve it for t.

46 Example 7 – Solution

47 Example 7 – Answer It will take the daughter 5/6 hours, or 50 minutes, to overtake her dad.


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