1 © 2010 Pearson Education, Inc. All rights reserved © 2010 Pearson Education, Inc. All rights reserved Chapter 1 Equations and Inequalities

OBJECTIVES © 2010 Pearson Education, Inc. All rights reserved 2 Applications of Linear Equations Learn procedures for solving applied problems. Use linear equations to solve applied problems. SECTION

Procedure for Solving Applied Problems Step 1Read the problem as many times as needed to understand it thoroughly. Pay close attention to the questions asked to help identify the quantity the variable should represent. Step 2Assign a variable to represent the quantity you are looking for, and, when necessary, express all other unknown quantities in terms of this variable. Frequently, it is helpful to draw a diagram to illustrate the problem or to set up a table to organize the information. 3 © 2010 Pearson Education, Inc. All rights reserved

Step 3Write an equation that describes the situation. Step 4Solve the equation. Step 5Answer the question asked in the problem. Step 6Check the answer against the description of the original problem (not just the equation solved in step 4). Procedure for Solving Applied Problems 4 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 1 Solving a Geometry Problem The length of a rectangle is 3 feet less than four times its width. Find the dimensions of the rectangle assuming that its perimeter is 64 feet. Solution Step 2Let x = the width of the rectangle. Then 4x – 3 = the length of the rectangle. 5 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 1 Solving a Geometry Problem Solution continued Step 3Perimeter = sum of the lengths of the four sides of the rectangle. Given 6 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 1 Solving a Geometry Problem Solution continued Step 4 Step 5 The width of the rectangle = 7 ft, and the length of the rectangle Step 6 Check: The perimeter is = 64 ft. 7 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 2 Analyzing Investments Tyrick invests $15,000, some in stocks and the rest in bonds. If he invests twice as much in stocks as he does in bonds, how much does he invest in each? Solution Step 2Let x = the amount invested in stocks. The rest of the $15,000 investment ($15,000 – x) is invested in bonds. We have one more important piece of information to use: 8 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 2 Analyzing Investments Solution continued Step 2 cont. Amount invested in stocks, x = Twice the amount invested in bonds, 15,000 – x Step 3 Replace the verbal description with algebraic expressions. Step 4 Distributive property 9 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 2 Analyzing Investments Solution continued Step 5Tyrick invests $10,000 in stocks and $15,000 – $10,000 = $5000 in bonds. Step 4 Distributive property Add 2x to both sides. Divide both sides by 3. Step 6Tyrick’s total investment is $10,000 + $5000 = $15,000, and $10,000 (stocks) is twice $5,000 (bonds). 10 © 2010 Pearson Education, Inc. All rights reserved

If a principal of P dollars is borrowed for a period of t years with interest rate r (expressed as a decimal) computed yearly, then the total interest paid at the end of years is Interest computed with this formula is called simple interest. When interest is computed yearly, the rate r is called an annual interest rate (or per annum interest rate). SIMPLE INTEREST 11 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 3 Solving Problems Involving Simple Interest Ms. Sharpy invests a total of $10,000 in blue- chip and technology stocks. At the end of a year, the blue-chips returned 12% and the technology stocks returned 8% on the original investments. How much was invested in each type of stock if the total interest earned was $1060? Solution We are asked to find two amounts: that invested in blue-chip stocks and that invested in technology stocks. Step 1 12 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 3 Solving Problems Involving Simple Interest Solution continued If we know how much was invested in blue-chip stocks, then we know that the rest of the $10,000 was invested in technology stocks. Let x = amount invested in blue-chip stocks. Then 10,000 – x = amount invested in technology stocks. Step 2 13 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 3 Solution continued InvestPrtI = Prt Bluex x Tech10,000 – x (10,000 – x) Interest from Blue-chip Interest from technology = Total Interest + 14 © 2010 Pearson Education, Inc. All rights reserved Solving Problems Involving Simple Interest

EXAMPLE 3 Solution continued Step 3 $ in blue-chip stocks Step 4 $ in technology stocks 15 © 2010 Pearson Education, Inc. All rights reserved Solving Problems Involving Simple Interest

EXAMPLE 3 Solution continued Step 5Ms. Sharpy invests $3500 in technology stocks and $6500 in blue- chip stocks. Step 6 16 © 2010 Pearson Education, Inc. All rights reserved Solving Problems Involving Simple Interest

If an object moves at a rate (average speed) r, then the distance traveled d in time t is UNIFORM MOTION 17 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 4 Solving a Uniform-Motion Problem A motorcycle police officer is chasing a car that is speeding at 70 miles per hour. The police officer is 3 miles behind the car and is traveling 80 miles per hour. How long will it be before the officer overtakes the car? Solution We are asked to find the amount of time before the officer overtakes the car. Draw a sketch to help visualize the problem. Step 2 18 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 4 Solving a Uniform-Motion Problem Solution continued Let x = distance in miles the car travels before being overtaken x + 3 = distance in miles the motorcycle travels before overtaking the car Step 2 19 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 4 Solving a Uniform-Motion Problem Solution continued The time from the start of the chase to the interception point is the same for both the car and the motorcycle. Object d miles r mph hours Carx70 Motorcyclex © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 4 Solving a Uniform-Motion Problem Solution continued The time required to overtake the car is Step 5 Step 3 Step 4 21 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 4 Solving a Uniform-Motion Problem Solution continued In of an hour, or 18 minutes, the police officer overtakes the car. Step 6Officer travels Car travels The officer started 3 miles behind the car so he does indeed overtake the car. 22 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 5 Dealing with a Bomb Threat on the QE II The Queen Elizabeth II was 1000 miles from Britain and traveling toward Britain at 32 miles per hour. A Hercules aircraft was flying from Britain directly toward the ship and averaging 300 miles per hour. How long would the passengers have to wait for the aircraft to meet the ship? Solution Step 1 The initial separation between the ship and the aircraft is 1000 miles. 23 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 5 Dealing with a Bomb Threat on the QE II Solution continued Step 2 Let t = time elapsed before meeting. 32t = distance ship traveled. 300t = distance aircraft traveled. 24 © 2010 Pearson Education, Inc. All rights reserved Distance ship traveled Distance aircraft traveled =1000 miles+ Step 3

EXAMPLE 5 Dealing with a Bomb Threat on the QE II Solution continued Step 5 The aircraft and ship meet after about 3 hours. 25 © 2010 Pearson Education, Inc. All rights reserved Step 4

The portion of a job completed per unit of time is called the rate of work. WORK RATE If a job can be completed in x units of time, then the portion of the job completed in one unit of time is. The portion of the job completed in t units of time is. When the portion of the job completed is 1, the job is done. 26 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 6 Solving a Work-Rate Problem One copy machine copies twice as fast as another. If both copiers work together, they can finish a particular job in 2 hours. How long would it take each copier, working alone, to do the job? Solution Step 1The speed of one copier is twice the speed of the other. We find out how long it takes the faster copier to do the job working alone. The slower copier will take twice as long. 27 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 6 Solving a Work-Rate Problem Solution continued Step 2 x = number of hours for the faster copier to complete the job alone. 2x = number of hours for the faster copier to complete the job alone. = portion of the job the faster copier does in 1 hour. 28 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 6 Solving a Work-Rate Problem Solution continued Portion done in 1 hr Time work together Potion done by each Faster copier 2 Slower copier 2 Slower copier portion Faster copier portion =1+ 29 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 6 Solving a Work-Rate Problem Solution continued Step 3 Step 4 Alone: faster copier, 3 hr; slower, 6 hrStep 5 Together = 1 job Faster copier, in 2 hr.Step 6 Slower copier, in 2 hr. 30 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 7 Solving a Mixture Problem A full 6-quart radiator contains 75% water and 25% pure antifreeze. How much of this mixture should be drained and replaced by pure antifreeze so that the resulting 6-quart mixture is 50% pure antifreeze? Solution Step 1Find the quantity of the radiator mixture that should be drained (25% pure antifreeze) and replaced by pure (100%) antifreeze. 31 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 7 Solving a Mixture Problem Solution continued Let x = quarts 25% antifreeze drained. x = quarts pure antifreeze added. 0.25x = quarts pure antifreeze drained. Step 2 Pure antifreeze final mix Pure antifreeze original mix Pure antifreeze drained Pure antifreeze added =–+ (50% of 6) = (25% of 6) – (25% of x) + x Step 3 32 © 2010 Pearson Education, Inc. All rights reserved

EXAMPLE 7 Solving a Mixture Problem Solution continued Step 3 Step 4 Step 5Drain 2 quarts of mixture from the radiator. Step 6Drain 2 qt from the 6, leaves 4 qt: 1 qt pure antifreeze, 3 qt water. Add 2 qt antifreeze, now have 3 qt antifreeze, 3 qt water, so it’s 50% pure antifreeze solution. 33 © 2010 Pearson Education, Inc. All rights reserved