§ 6.7 Formulas and Applications of Rational Equations

Applications of Rational Equations Formulas and mathematical models often contain rational expressions. You can often solve for a different variable by using the procedure for solving rational equations. The goal when solving for a specified variable is to get that variable alone on one side of the equation. To do so, collect all terms with this variable on one side and all other terms on the other side. Sometimes it is necessary to factor out the variable you are solving for. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 6.7

Applications of Rational Equations EXAMPLE Solve the formula for R: SOLUTION This is the original equation. Multiply both sides by the LCD, R + r. Simplify. Divide both sides by I and subtract r. Blitzer, Intermediate Algebra, 5e – Slide #3 Section 6.7

Applications of Rational Equations Check Point 1 Solve the formula for x: LCD= xyz Multiply both sides by the LCD, xyz. Subtract xz from both sides. Factor x and divide both sides by (y-z). Blitzer, Intermediate Algebra, 5e – Slide #4 Section 6.7

Applications of Rational Equations EXAMPLE A company is planning to manufacture small canoes. Fixed monthly cost will be $20,000 and it will cost $20 to produce each canoe. Write the cost function, C, of producing x canoes. Write the average cost function, , of producing x canoes. How many canoes must be produced each month for the company to have an average cost of $40 per canoe? Blitzer, Intermediate Algebra, 5e – Slide #5 Section 6.7

Applications of Rational Equations CONTINUED SOLUTION (a) The cost function, C, is the sum of the fixed cost and the variable costs. Fixed cost is $20,000. Variable cost: $20 for each canoe produced. (b) The average cost function, , is the sum of fixed and variable costs divided by the number of canoes produced. Blitzer, Intermediate Algebra, 5e – Slide #6 Section 6.7

Applications of Rational Equations CONTINUED (c) We are interested in the company’s production level that results in an average cost of $40 per canoe. Substitute 40, the average cost, for and solve the resulting rational equation for x. Substitute 40 for Multiply both sides by the LCD, x. Subtract 20x from both sides. Divide both sides by 20. The company must produce 1,000 canoes each month for an average cost of $40 per canoe. Blitzer, Intermediate Algebra, 5e – Slide #7 Section 6.7

Applications of Rational Equations Check Point 2 on page 454 A company is planning to manufacture wheelchairs that are light, fast, and beautiful. Fixed monthly cost will be $500,000 and it will cost $400 to produce each radically innovative chair. Write the cost function, C, of producing x wheelchairs. Write the average cost function, , of producing x wheelchairs. How many wheelchairs must be produced each month for the company to have an average cost of $450 per wheelchairs? Blitzer, Intermediate Algebra, 5e – Slide #8 Section 6.7

Applications of Rational Equations CONTINUED SOLUTION (a) The cost function, C, is the sum of the fixed cost and the variable costs. Fixed cost is $500,000. Variable cost: $400 for each wheelchair produced. (b) The average cost function, , is the sum of fixed and variable costs divided by the number of canoes produced. Blitzer, Intermediate Algebra, 5e – Slide #9 Section 6.7

Applications of Rational Equations CONTINUED (c) We are interested in the company’s production level that results in an average cost of $450 per wheelchair. Substitute 450, the average cost, for and solve the resulting rational equation for x. Substitute 450 for Multiply both sides by the LCD, x. Subtract 400x from both sides. Divide both sides by 50. The company must produce 10,000 wheelchairs each month for an average cost of $450 per wheelchairs. (Look at questions 15-18 on page 459.) Blitzer, Intermediate Algebra, 5e – Slide #10 Section 6.7

Applications of Rational Equations Time in Motion Blitzer, Intermediate Algebra, 5e – Slide #11 Section 6.7

Applications of Rational Equations EXAMPLE An engine pulls a train 140 miles. Then a second engine, whose average rate is 5 miles per hour faster than the first engine, takes over and pulls the train 200 miles. The total time required for both engines is 9 hours. Find the average rate of the first engine. SOLUTION 1) Let x represent one of the quantities. Let x = the rate of the first engine. 2) Represent other quantities in terms of x. Because the average rate of the second engine is 5 miles per hour faster than the average rate of the first engine, let x + 5 = the rate of the second engine. Blitzer, Intermediate Algebra, 5e – Slide #12 Section 6.7

Applications of Rational Equations CONTINUED 3) Write an equation that describes the conditions. By reading the problem again, we discover that the crucial idea is that the time for both engines’ trips is 9 hours. Thus, the time of the first engine plus the time of the second engine is 9 hours. Distance Rate Time Train 1 140 x Train 2 200 x + 5 The sum of the two times is 9 hours. The sum of the two times is 9 hours. The sum of the two times is 9 hours. We are now ready to write an equation that describes the problems’ conditions. Blitzer, Intermediate Algebra, 5e – Slide #13 Section 6.7

Applications of Rational Equations CONTINUED time of the second train plus Time of the first train equals 9 hours. 4) Solve the equation and answer the question. This is the equation for the problems’ conditions. Multiply both sides by the LCD, x(x + 5). Use the distributive property on both sides. Blitzer, Intermediate Algebra, 5e – Slide #14 Section 6.7

Applications of Rational Equations CONTINUED Simplify. Use the distributive property. Combine like terms. Subtract 340x + 700 from both sides. Factor the right side. Set each variable factor equal to zero. Solve for x. Blitzer, Intermediate Algebra, 5e – Slide #15 Section 6.7

Applications of Rational Equations CONTINUED Because x represents the average rate of the first engine, we reject the negative value, -20/9. The rate of the first engine is 35 miles per hour. 5) Check the proposed solution in the original wording of the problem. Do the two engines’ trips take a combined 9 hours? Because the rate of the second engine is 5 miles per hour faster than the rate of the first engine, the rate of the second engine is 35 + 5 = 40 miles per hour. The total time is 4 + 5 = 9 hours. This checks correctly. Blitzer, Intermediate Algebra, 5e – Slide #16 Section 6.7

Applications of Rational Equations EXAMPLE A plane travels 100 mi against the wind in the same time that it takes to travel 120 mi with the wind. The wind speed is 20 mph. Find the speed of the plane in still air. Let x = the speed of the plane in still air. Use d = rt, to complete the table D r t Against Wind 100 x – 20 With Wind 120 x + 20

Applications of Rational Equations continued Multiply by the LCD (x – 20)(x + 20). Cross multiply? The speed of the airplane is 220 mph in still air.

Applications of Rational Equations Check Point 3 on page 455 After riding at a steady speed for 40 miles, a bicyclist had a flat tire and walked 5 miles to a repair shop. The cycling rate was 4 times faster than the walking rate. If the time spent cycling and walking was 5 hours, at what rate was the cyclist riding? Let x = the rate walking. d r t Cycling 40 4x Walking 5 x

Applications of Rational Equations continued

Let x = the distance at reduced speed. Applications of Rational Equations EXAMPLE Dana Kenly drove 300 mi north from San Antonio, mostly on the freeway. She usually averaged 55 mph, but an accident slowed her speed through Dallas to 15 mph. If her trip took 6 hours, how many miles did she drive at the reduced speed? Let x = the distance at reduced speed. d r t Normal Speed (on freeway) 300 – x 55 Reduced Speed x 15 Total 6

Applications of Rational Equations continued Multiply by the LCD, 165. She drove 11 ¼ miles at the reduced speed. She drove 11 ¼ miles at reduced speed.

Applications of Rational Equations Work Problems In work problems, the number 1 represents one whole job completed. Equations in work problems are often based on the condition that the sum of the separate amount s of the job completed by each person (or machine) working on that job is equal to the whole job, or 1. Suppose there were only two people working on a job and that those two people completed the whole job. Then the following would be true: Fractional part of the job done by the first person fractional part of the job done by the second person 1 (one whole job completed). + = If it takes a person 10 hours to complete a job working alone, his rate is 1/10. If he ends up working x hours on that job, the fractional part of the job that he gets done is (x)(1/10) or x/10. Blitzer, Intermediate Algebra, 5e – Slide #23 Section 6.7

Applications of Rational Equations EXAMPLE A hurricane strikes and a rural area is without food or water. Three crews arrive. One can dispense needed supplies in 10 hours, a second in 15 hours, and a third in 20 hours. How long will it take all three crews working together to dispense food and water? SOLUTION 1) Let x represent one of the quantities. Let x = the time, in hours, for all three crews to do the job working together. 2) Represent other quantities in terms of x. There are no other unknown quantities. Blitzer, Intermediate Algebra, 5e – Slide #24 Section 6.7

Applications of Rational Equations CONTINUED 3) Write an equation that describes the conditions. Working together, the three crews can dispense the supplies in x hours. We construct a table to find the fractional part of the task completed by the three crews in x hours. Fractional part of job completed in 1 hour Time working together Fractional part of job completed in x hours First Crew 1/10 x x/10 Second Crew 1/15 x/15 Third Crew 1/20 x/20 Together =1/x =1 Blitzer, Intermediate Algebra, 5e – Slide #25 Section 6.7

Applications of Rational Equations CONTINUED Because all three teams working together can complete the job in x hours, 4) Solve the equation and answer the question. This is the equation for the problem’s conditions. Multiply both sides by 60, the LCD. 6 4 3 Use the distributive property on each side. Blitzer, Intermediate Algebra, 5e – Slide #26 Section 6.7

Applications of Rational Equations CONTINUED Simplify. Combine like terms. Divide both sides by 13. Because x represents the time that it would take all three crews to get the job done working together, the three crews can get the job done in about 4.6 hours. 5) Check the proposed solution in the original wording of the problem. Will the three crews complete the job in 4.6 hours? Because the first crew can complete the job in 10 hours, in 4.6 hours, they can complete 4.6/10, or 0.46, of the job. Blitzer, Intermediate Algebra, 5e – Slide #27 Section 6.7

Applications of Rational Equations CONTINUED Because the second crew can complete the job in 15 hours, in 4.6 hours, they can complete 4.6/15, or 0.31, of the job. Because the third crew can complete the job in 20 hours, in 4.6 hours, they can complete 4.6/20, or 0.23, of the job. Notice that 0.46 + 0.31 + 0.23 = 1, which represents the completion of the entire job, or one whole job. Blitzer, Intermediate Algebra, 5e – Slide #28 Section 6.7

Fractional Part of the Job Done Applications of Rational Equations EXAMPLE Stan needs 45 minutes to do the dishes, while Bobbie can do them in 30 minutes. How long will it take them if they work together? Let x = the time it will take them working together. Rate Time Working Together Fractional Part of the Job Done Stan Bobbie Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Applications of Rational Equations continued Part done Part done 1 whole by Stan plus by Bobbie equals job. Multiply by the LCD, 90. It will take them 18 minutes working together.

Fractional Part of the Job Done Applications of Rational Equations Check Point 4 on page 457 A new underwater tunnel is being built using tunnel-boring machines that begin at opposite ends of the tunnel. One tunnel-boring machine can complete the tunnel in 18 months. A faster machine can tunnel to the other side in 9 months. If both machines start at opposite ends and work at the same time, in how many months will the tunnel be finished? Rate Time Working Together Fractional Part of the Job Done Slower machine Faster machine Blitzer, Intermediate Algebra, 5e – Slide #31 Section 6.7

Applications of Rational Equations CONTINUED Solve the equation and answer the question. This is the equation Multiply both sides by 18, the LCD. 2 Use the distributive property on each side. Simplify. Combine like terms. Divide both sides by 3. Blitzer, Intermediate Algebra, 5e – Slide #32 Section 6.7

Applications of Rational Equations Check Point 5 An experienced carpenter can panel a room 3 times faster than an apprentice can. Working together, they can panel the room in 6 hours. How long would it take each person working alone to do the job? Fractional part of job completed in 1 hour Time Working Together Fractional Part of the Job Completed in 6 hours Experienced carpenter Apprentice Blitzer, Intermediate Algebra, 5e – Slide #33 Section 6.7

Applications of Rational Equations CONTINUED Solve the equation and answer the question. This is the equation Multiply both sides by 3x, the LCD. Use the distributive property on each side. Simplify. Combine like terms. Divide both sides by 3. Blitzer, Intermediate Algebra, 5e – Slide #34 Section 6.7

Applications of Rational Equations Do 48 on page 461 A pool can be filled by a pipe in 3 hours. It akes 3 times as long for another pipe to empty the pool. How long will it take to fill the pool if both pipes are open? Blitzer, Intermediate Algebra, 5e – Slide #35 Section 6.7

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