Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 1 Equations and Inequalities Copyright © 2013, 2009, 2005 Pearson Education, Inc.

2 Applications and Modeling with Linear Equations 1.2 Solving Applied Problems Geometry Problems Motion Problems Mixture Problems Modeling with Linear Equations

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 3 SOLVING AN APPLIED PROBLEM Step 1 Read the problem carefully until you understand what is given and what is to be found. Step 2 Assign a variable to represent the unknown value, using diagrams or tables as needed. Write down what the variable represents. If necessary, express any other unknown values in terms of the variable.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 4 SOLVING AN APPLIED PROBLEM Step 3 Write an equation using the variable expression(s). Step 4 Solve the equation. Step 5 State the answer to the problem. Does it seem reasonable? Step 6 Check the answer in the words of the original problem.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 5 EXAMPLE 1 FINDING THE DIMENSIONS OF A SQUARE If the length of each side of a square is increased by 3 cm, the perimeter of the new square is 40 cm more than twice the length of the original square. Find the dimensions of the original square. Solution Step 1 Read the problem. We must find the length of the original square.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 6 EXAMPLE 1 FINDING THE DIMENSIONS OF A SQUARE Step 2 Assign a variable. Since the length of a side of the original square is to be found, let the variable represent this length. The length of a side of the new square is 3 cm more than the length of a side of the old square. x x x + 3 Original square Side is increased by 3.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 7 EXAMPLE 1 FINDING THE DIMENSIONS OF A SQUARE Step 2 Assign a variable. Write a variable expression for the perimeter of the new square. The perimeter of a square is 4 times the length of a side.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 8 EXAMPLE 1 FINDING THE DIMENSIONS OF A SQUARE Step 3 Write an equation. Translate the English sentence that follows into its equivalent algebraic expression. The new perimeter is 40 more than twice the length of a side of the original square.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 9 EXAMPLE 1 FINDING THE DIMENSIONS OF A SQUARE Step 4 Solve the equation. Distributive property Subtract 2x and 12. Divide by 2.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 10 EXAMPLE 1 FINDING THE DIMENSIONS OF A SQUARE Step 5 State the answer. Each side of the original square measures 14 cm.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 11 EXAMPLE 1 FINDING THE DIMENSIONS OF A SQUARE Step 6 Check. Go back to the words of the original problem to see that all necessary conditions are satisfied. The length of a side of the new square would be = 17 cm. The perimeter of the new square would be 4(17)= 68 cm. Twice the length of a side of the original square would be 2(14) = 28 cm. Since = 68, the answer checks.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 12 Motion Problems Problem-Solving Hint In a motion problem, the components distance, rate, and time are denoted by the letters d, r, and t, respectively. (The rate is also called the speed or velocity. Here, rate is understood to be constant.) and its related formsand

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 13 Example 2 SOLVING A MOTION PROBLEM Maria and Eduardo are traveling to a business conference. The trip takes 2 hr for Maria and 2.5 hr for Eduardo, since he lives 40 mi farther away. Eduardo travels 5 mph faster than Maria. Find their average rates.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 14 Example 2 SOLVING A MOTION PROBLEM Step 1 Read the problem. We must find Maria’s and Eduardo’s average rates. Solution

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 15 Example 2 SOLVING A MOTION PROBLEM Step 2 Assign a variable. Since average rates are to be found, we let the variables represent one of these rates. Let x = Maria’s rate. Then x + 5 = Eduardo’s rate. Summarize the given information in a table. rtd Mariax22x2x Eduardox (x + 5) Use d = rt.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 16 Example 2 SOLVING A MOTION PROBLEM Step 3 Write an equation. Eduardo’s distance traveled exceeds Maria’s distance by 40 mi. Translate this relationship into its algebraic form. Eduardo’s distance is 40 more than Maria’s.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 17 Example 2 SOLVING A MOTION PROBLEM Step 4 Solve. Distributive property Subtract 2x and Divide by 0.5

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 18 Example 2 SOLVING A MOTION PROBLEM Step 5 State the answer. Maria’s rate of travel is 55 mph, and Eduardo’s rate is = 60 mph. Step 6 Check. The diagram shows that the conditions of the problem are satisfied. Distance traveled by Maria: 2(55) = 110 mi Distance traveled by Eduardo: 2.5(60) = 150 mi 150 – 110 = 40

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 19 Mixture Problems Problem-Solving Hint In mixture problems involving solutions, the rate (percent) of concentration is multiplied by the quantity to get the amount of pure substance present. The concentration of the final mixture must be between the concentrations of the two solutions making up the mixture.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 20 Example 3 SOLVING A MIXTURE PROBLEM Lisa Harmon is a chemist. She needs a 20% solution of alcohol. She has a 15% solution on hand, as well as a 30% solution. How many liters of the 15% solution should she add to 3L of the 30% solution to obtain her 20% solution?

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 21 Example 3 SOLVING A MIXTURE PROBLEM Solution Step 1 Read the problem. We must find the required number of liters of 15% alcohol solution.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 22 Example 3 SOLVING A MIXTURE PROBLEM Step 2 Assign a variable. Let x = the number of liters of 15% solution to be added. StrengthLiters of Solution Liters of Pure Alcohol 15%x0.15x 30%30.30(3) 20%x (x + 3) Sum must equal

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 23 Example 3 SOLVING A MIXTURE PROBLEM Step 3 Write an equation. The number of liters of pure alcohol in the 15% solution plus the number of liters in the 30% solution must equal the number of liters in the final 20% solution. Liters of pure alcohol in 15% = Liters of pure alcohol in 20% + Liters of pure alcohol in 30%

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 24 Example 3 SOLVING A MIXTURE PROBLEM Step 4 Solve. Distributive property Subtract 0.60 and 0.15x. Divide by 0.05.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 25 Example 3 SOLVING A MIXTURE PROBLEM Step 5 State the answer. Thus, 6 L of 15% solution should be mixed with 3 L of 30% solution, giving = 9 L of 20% solution. Step 6 Check. The answer checks since the amount of alcohol in the two solutions is equal to the amount of alcohol in the mixture. Solutions Mixture

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 26 Mixture Problems Problem-Solving Hint In mixed investment problems, multiply each principal by the interest rate and the time in years to find the amount of interest earned.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 27 Example 4 SOLVING AN INVESTMENT PROBLEM An artist has sold a painting for $410,000. He needs some of the money in 6 months and the rest in 1 yr. He can get a Treasury bond for 6 months at 2.65% and one for a year at 2.91%. His broker tells him the two investments will earn a total of $8761. How much should be invested at each rate to obtain that amount of interest?

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 28 Example 4 SOLVING AN INVESTMENT PROBLEM Step 1 Read the problem. We must find the amount to be invested at each rate. Solution

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 29 Example 4 SOLVING AN INVESTMENT PROBLEM Step 2 Assign a variable. Let x = dollar amount invested for 6 months at 2.65%. 410,000 – x = dollar amount invested for 1 yr at 2.91%. Invested AmountInterest Rate (%) Time (in years) Interest Earned x x(0.0265)(0.5) 410,000 – x 2.911(410,000 – x)(0.0291)(1)

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 30 Example 4 SOLVING AN INVESTMENT PROBLEM Step 3 Write an equation. The sum of the two interest amounts must equal the total interest earned. Interest from 2.65% investment = Total interest + Interest from 2.91% investment

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 31 Example 4 SOLVING AN INVESTMENT PROBLEM Step 4 Solve.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 32 Example 4 SOLVING AN INVESTMENT PROBLEM Step 5 State the answer. The artist should invest $200,000 at 2.65% for 6 months and $410,000 – $200,000 = $210,000 at 2.91% for 1 yr to earn $8761 in interest.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 33 Example 4 SOLVING AN INVESTMENT PROBLEM Step 6 Check. The 6-month investment earns while the 1-yr investment earns The total amount of interest earned is

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 34 Modeling with Linear Equations A mathematical model is an equation (or inequality) that describes the relationship between two quantities. A linear model is a linear equation.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 35 Example 5 MODELING THE PREVENTION OF INDOOR POLLUTANTS If a vented range hood removes contaminants such as carbon monoxide and nitrogen dioxide from the air at a rate of F liters of air per second, then the percent P of contaminants that are also removed from the surrounding air can be modeled by the linear equation where What flow F must a range hood have to remove 50% of the contaminants from the air? (Source: Proceedings of the Third International Conference on Indoor Air Quality and Climate.)

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 36 Example 5 MODELING THE PREVENTION OF INDOOR POLLUTANTS Solution Replace P with 50 in the linear model, and solve for F. Subtract Divide by Therefore, to remove 50% of the contaminants, the flow rate must be approximately L of air per second. Let P = 50. Given model

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 37 Example 6 MODELING HEALTH CARE COSTS The projected per capita health care expenditures in the United States, where y is in dollars, and x is years after 2000, are given by the linear equation. (Source: Centers for Medicare and Medicaid Services.) (a) What were the per capita health care expenditures in the year 2010? (b) If this model continues to describe health care expenditures, when will the per capita expenditures reach $9200? Linear model

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 38 Example 6 MODELING HEALTH CARE COSTS In part (a) we are given information to determine a value for x and asked to find the corresponding value of y, whereas in part (b) we are given a value for y and asked to find the corresponding value of x. (a) The year 2010 is 10 yr after the year Let x = 10 and find the value of y. In 2010, the estimated per capita health care expenditures were $7942. Given model Let x = 10. Multiply and then add.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 39 Example 6 MODELING HEALTH CARE COSTS (b) Let y = 9200 in the given model, and find the value of x. The x-value of 13.7 indicates that per capita health care expenditures are projected to reach $9200 during the 13th year after 2000—that is, corresponds to = Let y = Subtract Divide by 343.