Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 1

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 2 Systems of Linear Equations and Inequalities Chapter 4

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide Applications of Linear Systems

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 4 Objectives 1.Solve problems about unknown numbers. 2.Solve problems about quantities and their costs. 3.Solve problems about mixtures. 4.Solve problems about distance, rate (or speed), and time. 4.4 Applications of Linear Systems

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 5 Solving an Applied Problem with Two Variables Step 1Read the problem, several times if necessary, until you understand what is given and what is to be found. Step 2Assign variables to represent the unknown values, using diagrams or tables as needed. Write down what each variable represents. Step 3Write two equations using both variables. Step 4Solve the system of two equations. Step 5State the answer to the problem. Is the answer reasonable? Step 6 Check the answer in the words of the original problem. 4.4 Applications of Linear Systems

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 6 Example 1 Solving Problems about Unknown Numbers 4.4 Applications of Linear Systems Amy and Bryan just celebrated a birthday together. Amy is 3 years older than Bryan. However, if Bryan doubled his age and subtracted 5, the result would be 4 years older than Amy. How old are Amy and Bryan? Step 1 Read the problem carefully. We are looking for Amy and Bryan’s ages. Step 2 Assign variables. Let x = Amy’s age, and y = Bryan’s age. Step 3 Write two equations. Since Amy is 3 years older than Bryan, x = 3 + y. And, since Bryan’s age doubled minus 5 is 4 years more than Amy’s age, 2y – 5 = 4 + x.

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 7 Example 1 (continued) Solving Problems about Unknown Numbers 4.4 Applications of Linear Systems Step 4 Solve. x = 3 + y 2y – 5 = 4 + x x – y = 3 –x + 2y = 9 + y = 12 Step 5 State the answer. We will substitute y = 12 into the first equation to solve for x. x – y = 3 x – 12 = x = 15 Amy is 15 years old, and Bryan is 12 years old.

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 8 Example 1 (concluded) Solving Problems about Unknown Numbers 4.4 Applications of Linear Systems Step 6 Check. x – y = 3 2y – 5 = 4 + x 15 – 12 = 3 ? 3 = 3 2(12) – 5 = ? 24 – 5 = 19 ? 19 = 19 CAUTION If an applied problem asks for two values, be sure to give both of them in your answer.

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 9 Example 2 Solving Problems about Quantities and Their Costs 4.4 Applications of Linear Systems Carrie and Diego are headed to the vending machine and have decided to pool their money. Carrie has only quarters, Diego has only dimes, and together they have ten coins worth $2.05. How many quarters does Carrie have, and how many dimes does Diego have? Step 1 Read. We are looking for the number of quarters for Carrie and the number of dimes for Diego. Step 2 Assign variables. # of CoinsCoin ValueTotal Value Carriex x Diegoy y Totalx + y0.25x y

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 10 Example 2 (continued) 4.4 Applications of Linear Systems Step 4 Solve. x + y = x y = 2.05 –0.10x – 0.10y = –1 0.25x y = x = 1.05 Solving Problems about Quantities and Their Costs y = y = 10 –7 x = 7 Carrie had 7 quarters and Diego had 3 dimes. Step 5 State the Answer. We will substitute x = 7 into the first equation to solve for y. Step 3 Write two equations. Since they had 10 coins together, x + y = 10. And, since their coins were worth $2.05, 0.25x y = –0.10( )( )·–0.10

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 11 Example 2 (concluded) 4.4 Applications of Linear Systems Step 6 Check = 10 ? 10 = (7) (3) = 2.05 ? = 2.05 ? 2.05 = 2.05 Solving Problems about Quantities and Their Costs x + y = x y = 2.05

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 12 Example 3 Solving Problems about Mixtures 4.4 Applications of Linear Systems A pharmacist needs 100L of 50% alcohol solution. She has on hand 30% alcohol solution and 80% alcohol solution, which she can mix. How many liters of each will be required to make the 100L of 50% alcohol solution? Step 1 Read. We are looking for the number of liters of 30% and 80% alcohol solutions to get a 50% alcohol solution. Step 2 Assign Variables. Percent# of LitersLiters of Alcohol 0.30x0.30x 0.80y0.80y (100)

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 13 Example 3 (continued) 4.4 Applications of Linear Systems Step 4 Solve. x + y = x y = 50 –0.30x – 0.30y = – x y = y = 20 Solving Problems about Mixtures x = 60 x + 40 = 100 –40 y = 40 The pharmacist needs 60L of 30% alcohol solution and 40L of 80% alcohol solution. Step 5 State the answer. We substitute y = 40 into the first equation to solve for x. Step 3 Write two equations. Since we need 100L of solution, x + y = 100. Since it must be 50% alcohol, 0.30x y = 0.50(100). –0.30( )( )·–0.30

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 14 Example 3 (concluded) 4.4 Applications of Linear Systems Step 6 Check = 100 ? 100 = (60) (40) = 50 ? = 50 ? 50 = 50 Solving Problems about Mixtures x + y = x y = 50

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 15 Example 4 Solving Problems about Distance, Rate (or Speed), and Time 4.4 Applications of Linear Systems A plane flies 560 mi in 1.75 hr traveling with the wind. The return trip against the same wind takes the plane 2hr. Find the speed of the plane and the speed of the wind. Step 1 Read. We are looking for the speed of the plane and the speed of the wind. Step 2 Assign variables. Let x = speed of the plane and y = speed of the wind. rtd With Wind x + y Against Wind x – y2560

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 16 2( ) 1.75x y = 560 2x – 2y = 560 Example 4 (continued) 4.4 Applications of Linear Systems Step 4 Solve. 3.50x y = x – 3.50y = x = 2100 –2y = –40 2(300) – 2y = 560 –600 x = 300 The speed of the plane was 300 mph and the speed of the wind was 20 mph. Step 5 State the answer. We will substitute x = 300 into the second equation to solve for y. Step 3 Write two equations. The trip with the wind is (x + y)1.75 = 560. The trip against the wind is (x – y)2 = 560. ( )2 Solving Problems about Distance, Rate (or Speed), and Time 1.75( )( ) – 2y = 560 y = 20

Copyright © 2010 Pearson Education, Inc. All rights reserved. 4.4 – Slide 17 Example 4 (concluded) 4.4 Applications of Linear Systems Step 6 Check. 1.75(300) (20) = 560 ? 560 = 560 2(300) – 2(20) = 560 ? 600 – 40 = 560 ? 560 = x y = 560 2x – 2y = = 560 ? Solving Problems about Distance, Rate (or Speed), and Time