1. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 14.4 Systems of Linear Equations and Problem Solving

2. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 22 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Problem-Solving Steps 1.UNDERSTAND the problem. During this step, become comfortable with the problem. Some ways of doing this are to Read and reread the problem. Choose two variables to represent the two unknowns. Construct a drawing. Propose a solution and check Pay careful attention to check your proposed solution. This will help when writing equations to model the problem. Problem Solving Steps Continued

3. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 33 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 2.TRANSLATE the problem into two equations. 3.SOLVE the system of equations. 4.INTERPRET the results. Check the proposed solution in the stated problem and state your conclusion. Problem Solving Steps

4. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 44 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Finding an Unknown Number Example Continued One number is 4 more than twice the second number. Their total is 25. Find the numbers. Read and reread the problem. Suppose that the second number is 5. Then the first number, which is 4 more than twice the second number, would have to be 14 (4 + 25). Is their total 25? No: 14 + 5 = 19. Our proposed solution is incorrect, but we now have a better understanding of the problem. Since we are looking for two numbers, we let x = first number y = second number 1.UNDERSTAND

5. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 55 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Continued 2. TRANSLATE Example continued Finding an Unknown Number One number is 4 more than twice the second number. x = 4 + 2y Their total is 25. x + y = 25

6. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 66 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Example continued 3.SOLVE Continued Finding an Unknown Number Using the substitution method, we substitute the solution for x from the first equation into the second equation. x + y = 25 (4 + 2y) + y = 25 Replace x with 4+2y. 4 + 3y = 25 Simplify. 3y = 21 Subtract 4 from both sides. y = 7 Divide both sides by 3. We are solving the system x = 4 + 2y x + y = 25

7. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 77 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Example continued 4.INTERPRET Finding an Unknown Number Check: Substitute x = 18 and y = 7 into both of the equations. First equation: x = 4 + 2y 18 = 4 + 2(7) True Second equation: x + y = 25 18 + 7 = 25 True State: The two numbers are 18 and 7. Now we substitute 7 for y into the first equation. x = 4 + 2y = 4 + 2(7) = 4 + 14 = 18

8. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 88 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving a Problem About Prices Example Continued Hilton University Drama club sold 311 tickets for a play. Student tickets cost 50 cents each; non-student tickets cost $1.50. If the total receipts were $385.50, find how many tickets of each type were sold. 1.UNDERSTAND Read and reread the problem. Suppose the number of students tickets was 200. Since the total number of tickets sold was 311, the number of non-student tickets would have to be 111 (311 – 200).

9. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 99 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Continued The total receipts are $385.50. Admission for the 200 students will be 200($0.50), or $100. Admission for the 111 non- students will be 111($1.50) = $166.50. This gives total receipts of $100 + $166.50 = $266.50. Our proposed solution is incorrect, but we now have a better understanding of the problem. Since we are looking for two numbers, we let s = the number of student tickets n = the number of non-student tickets Example continued 1.UNDERSTAND (continued) Solving a Problem About Prices

10. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 10 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Continued 2.TRANSLATE Hilton University Drama club sold 311 tickets for a play. s + n = 311 total receipts were $385.50 0.50s Total receipts = 385.50 Admission for students 1.50n Admission for non-students + Example continued Solving a Problem About Prices

11. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 11 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Example continued 3.SOLVE Continued We are solving the system s + n = 311 0.50s + 1.50n = 385.50 Since the equations are written in standard form (and we might like to get rid of the decimals anyway), we’ll solve by the addition method. Multiply the second equation by –2. s + n = 311 –2(0.50s + 1.50n) = –2(385.50) s + n = 311 –s – 3n = –771 –2n = –460 n = 230 Solving a Problem About Prices

12. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 12 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Example continued 4.INTERPRET Check: Substitute s = 81 and n = 230 into both of the equations. s + n = 311 First Equation 81 + 230 = 311 True 0.50s + 1.50n = 385.50 Second Equation 40.50 + 345 = 385.50 True 0.50(81) + 1.50(230) = 385.50 State: There were 81 student tickets and 230 non student tickets sold. Now we substitute 230 for n into the first equation to solve for s. s + n = 311 s + 230 = 311 s = 81 Solving a Problem About Prices

13. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 13 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving a Rate Problem Example Continued Terry Watkins can row about 10.6 kilometers in 1 hour downstream and 6.8 kilometers upstream in 1 hour. Find how fast he can row in still water, and find the speed of the current. 1.UNDERSTAND Read and reread the problem. We are going to propose a solution, but first we need to understand the formulas we will be using. Although the basic formula is d = r t (or r t = d), we have the effect of the water current in this problem. The rate when traveling downstream would actually be r + w and the rate upstream would be r – w, where r is the speed of the rower in still water, and w is the speed of the water current.

14. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 14 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving a Rate Problem Example continued 1.UNDERSTAND (continued) Suppose Terry can row 9 km/hr in still water, and the water current is 2 km/hr. Since he rows for 1 hour in each direction, downstream would be (r + w)t = d or (9 + 2)1 = 11 km Upstream would be (r – w)t = d or (9 – 2)1 = 7 km Our proposed solution is incorrect (hey, we were pretty close for a guess out of the blue), but we now have a better understanding of the problem. Since we are looking for two rates, we let r = the rate of the rower in still water w = the rate of the water current Continued

15. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 15 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 2.TRANSLATE Example continued Solving a Rate Problem rate downstream (r + w) time downstream 1 distance downstream = 10.6 rate upstream (r – w) time upstream 1 distance upstream = 6.8 Continued

16. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 16 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Example continued 3.SOLVE Continued Solving a Rate Problem We are solving the system r + w = 10.6 r – w = 6.8 Since the equations are written in standard form, we’ll solve by the addition method. Simply add the two equations together. r + w = 10.6 r – w = 6.8 2r = 17.4 r = 8.7

17. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 17 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Example continued 4.INTERPRET Solving a Rate Problem Check: Substitute r = 8.7 and w = 1.9 into both equations. (r + w)1 = 10.6 First equation (8.7 + 1.9)1 = 10.6 True (r – w)1 = 1.9 Second equation (8.7 – 1.9)1 = 6.8 True State: Terry’s rate in still water is 8.7 km/hr and the rate of the water current is 1.9 km/hr. Now we substitute 8.7 for r into the first equation. r + w = 10.6 8.7 + w = 10.6 w = 1.9

18. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 18 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving a Mixture Problem Example Continued A Candy Barrel shop manager mixes M&M’s worth $2.00 per pound with trail mix worth $1.50 per pound. How many pounds of each should she use to get 50 pounds of a party mix worth $1.80 per pound? 1.UNDERSTAND Read and reread the problem. We are going to propose a solution, but first we need to understand the formulas we will be using. To find out the cost of any quantity of items we use the formula price per unitnumber of units=price of all units

19. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 19 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving a Mixture Problem Example continued Continued 1.UNDERSTAND (continued) Suppose the manage decides to mix 20 pounds of M&M’s. Since the total mixture will be 50 pounds, we need 50 – 20 = 30 pounds of the trail mix. Substituting each portion of the mix into the formula, M&M’s $2.00 per lb 20 lbs = $40.00 trail mix $1.50 per lb 30 lbs = $45.00 Mixture $1.80 per lb 50 lbs = $90.00

20. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 20 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving a Mixture Problem Continued 1. UNDERSTAND (continued) Since $40.00 + $45.00 ≠ $90.00, our proposed solution is incorrect (hey, we were pretty close again), but we now have a better understanding of the problem. Since we are looking for two quantities, we let x = the amount of M&M’s y = the amount of trail mix Example continued

21. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 21 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Continued 2. TRANSLATE Example continued Solving a Mixture Problem Fifty pounds of party mix x + y = 50 price per unit number of units = price of all units Using Price of M&M’s 2x2x Price of trail mix + 1.5y Price of mixture = 1.8(50) = 90

22. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 22 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Example continued 3. SOLVE Continued Solving a Mixture Problem We are solving the system x + y = 50 2x + 1.50y = 90 Since the equations are written in standard form, we’ll solve by the addition method. Multiply the first equation by 3 and the second equation by –2 (which will also get rid of the decimal). 3(x + y) = 3(50) –2(2x + 1.50y) = –2(90) 3x + 3y = 150 –4x – 3y = –180 –x = –30 x = 30

23. Martin-Gay, Prealgebra & Introductory Algebra, 3ed 23 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Example continued 4.INTERPRET Solving a Mixture Problem Check: Substitute x = 30 and y = 20 into both of the equations. x + y = 50 First equation 30 + 20 = 50 True 2x + 1.50y = 90 Second equation 60 + 30 = 90 True 2(30) + 1.50(20) = 90 State: The store manager needs to mix 30 pounds of M&M’s and 20 pounds of trail mix to get the mixture at $1.80 a pound. Now we substitute 30 for x into the first equation. x + y = 50 30 + y = 50 y = 20