Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solving Applications Using Rational Equations Problems Involving Work Problems Involving Motion 6.5

Slide 6- 2 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Problems Involving Work Example Matt and Adam need to repaint their parent’s house. Matt can do the job alone in 45 hours and Adam can do the job alone in 60 hours. How long will it take the two of them, working together, to paint the house?

Slide 6- 3 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution 1.Familiarize. We familiarize ourselves with the problem by considering two incorrect ways of translating the problem to mathematical language. a) One incorrect way to translate the problem is to add the two times: 45 hr + 60 hr = 105 hr. Think about this. Matt can do the job alone in 45 hr. If Matt and Adam work together, whatever time it takes them must be less than 45 hr.

Slide 6- 4 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley b) Another incorrect approach is to assume that each person paints half of the house. Were this the case, Matt would paint ½ the house in ½(45 hr), or 22.5 hr, and Adam would paint ½ the house in ½(60 hr) or 30 hr. But time would be wasted since Matt would finish 7.5 hr before Adam. Were Matt to help Adam after completing his half, the entire job would take between 22.5 and 30 hr. Solution (continued)

Slide 6- 5 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley We are looking for the time required to paint 1 entire house, not just part of it. To find this time, we form a table and extend the pattern. Fractionof theHouse Painted TimeBy MattBy AdamTogether 1 hr 2 hr 3 hr t hr Solution (continued)

Slide 6- 6 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Translate. From the table, we see that t must be some number for which Fraction of house done by Matt in t hr Fraction of house done by Adam in t hr or 3. Carry out. We solve the equation: Multiplying by the LCD Solution (continued)

Slide 6- 7 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Distributing the 180 and simplifying 4. Check. In 180/7 hr, Matt paints 1/45(180/7), or 4/7, of the house and Adam paints 1/60(180/7), or 3/7, of the house. Together, they paint 4/7 + 3/7, or 1 house. The fact that our solution is between 22.5 and 30 hr is also a check. Solution (continued) 5. State. It will take 25 and 5/7 hr for Matt and Adam, working together, to pain the house.

Slide 6- 8 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Modeling Work Problems If a = the time needed for A to complete the work alone b = the time needed for B to complete the work alone t = the time needed for A and B to complete the work together, then The following are equivalent equations that can also be used:

Slide 6- 9 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Problems Involving Motion Problems dealing with distance, rate (or speed), and time are called motion problems. To translate them, we use either the basic motion formula, d = rt, or the formulas r = d/t or t = d/r, which can be derived from d = rt.

Slide Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Example Solution Jake’s new scooter goes 4 mph faster than Josh does on his scooter. In the time it takes Jake to travel 54 miles, Josh travels 48 miles. Find the speed of each scooter. 1. Familiarize. Let’s guess that Jake is going 20 mph. Josh would then be traveling 20 – 4, or 16 mph. At 16 mph, he would travel 48 miles in 3 hr. Going 20 mph, Jake would cover 54 mi in 54/20 = 2.7 hr. Since 3  2.7, our guess was wrong, but we can see that if r = the rate, in miles per hour, of Jake’s scooter, then the rate of Josh’s scooter = r – 4.

Slide Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Make a drawing and constructing a table can be helpful. DistanceSpeedTime Jake’s Scooter Josh’s Scooter Solution (continued)

Slide Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 2. Translate. By looking at how we checked our guess, we see that in the Time column of the table, the t’s can be replaced, using the formula Time = Distance/Rate, as follows. DistanceSpeedTime Jake’s Scooter Josh’s Scooter Since we are told that the times must be the same, we can write an equation: Solution (continued)

Slide Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3. Carry out. We solve the equation: Solution (continued)

Slide Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4. Check. If our answer checks, Jake’s scooter is going 36 mph and Josh’s scooter is going 36 – 4 = 32 mph. Traveling 54 miles at 36 mph, Jake is riding for 54/36 or 1.5 hours. Traveling 48 miles at 32 mph, Josh is riding for 48/32 or 1.5 hours. Our answer checks since the two times are the same. 5. State. Jake’s speed is 36 mph, and Josh’s speed is 32 mph. Solution (continued)