1. Copyright © Cengage Learning. All rights reserved. Equations and Inequalities 2

2. Copyright © Cengage Learning. All rights reserved. Section 2.6 Motion and Mixture Applications

3. 3 Objectives 1. Solve a motion application using a linear equation in one variable. 2. Solve a liquid mixture application using a linear equation in one variable. 3. Solve a dry mixture application using a linear equation in one variable. 1 1 2 2 3 3

4. 4 Motion and Mixture Applications In this section, we consider uniform motion and mixture applications. In these problems, we will use the following three formulas: r  t = d r  b = a v = p  n The rate multiplied by the time equals the distance. The rate multiplied by the base equals the amount. The value equals the price multiplied by the number.

5. 5 Solve a motion application using a linear equation in one variable 1.

6. 6 Example – Traveling Chicago and Green Bay are about 200 miles apart. If a car leaves Chicago traveling toward Green Bay at 55 mph at the same time as a truck leaves Green Bay bound for Chicago at 45 mph, how long will it take them to meet? 1.What am asked to find? Time it takes for the two vehicles to meet: t.

7. 7 Example – Traveling 2.Form an equation Motion problems are based on the formula d = rt, where d is the distance traveled, r is the rate, and t is the time. cont’d Figure 2-11

8. 8 Example – Traveling We know that the two vehicles travel for the same amount of time, t hours. When the they meet, the total distance traveled is 200 miles. 55t + 45t = 200 cont’d

9. 9 Example – Traveling 3.Solve the equation 55t + 45t = 200 100t = 200 t = 2 cont’d

10. 10 Example – Traveling 4.Check the result In 2 hours, the car will travel 55  2 = 110 miles, while the truck will travel 45  2 = 90 miles. The total distance traveled will be 110 + 90 = 200 miles. cont’d

11. 11 Solve a liquid mixture application using a linear equation in one variable 2.

12. 12 Example – Mixing Acid A chemist has one solution that is 50% sulfuric acid and another that is 20% sulfuric acid. How much of each should she use to make 12 liters of a solution that is 30% sulfuric acid? 1.What am asked to find? Number of liters for the 50% sulfuric acid solution: x Number of liters for the 20% sulfuric acid soluton: 12 - x

13. 13 Example – Mixing Acid 2.Form an equation Liquid mixture applications are based on the relationship rb = a, where b is the base, r is the rate, and a is the amount. If x represents the number of liters of 50% solution to use, the amount of sulfuric acid in the solution will be 0.50x liters. The amount of sulfuric acid in the 20% solution will be 0.20(12 – x) liters. The amount of sulfuric acid in the final mixture will be 0.30(12) liters. cont’d

14. 14 Example – Mixing Acid We can organize information in a chart or a diagram, as shown in Figure 2-14. (a) (b) Figure 2-14 cont’d

15. 15 Example – Mixing Acid Since the number of liters of sulfuric acid in the 50% solution plus the number of liters of sulfuric acid in the 20% solution will equal the number of liters of sulfuric acid in the mixture, we can form the equation: cont’d

16. 16 Example Mixing Acid 3.Solve the equation 0.5x + 0.2(12 – x) = 0.3(12) 5x + 2(12 – x) = 3(12) 5x + 24 – 2x = 36 3x + 24 = 36 3x = 12 x = 4 (liters of 50% acid) 12 – 4 = 8 (liters of 20% acid) cont’d

17. 17 Example – Mixing Acid State the conclusion Recall that we let x represent the number of liters of the 50% sulfuric acid solution. The chemist must mix 4 liters of the 50% solution and 12 – 4 = 8 liters of the 20% solution. Check the result The amount of acid in 4 liters of 50% solution is 4(0.50) = 2 liters. The amount of acid in 8 liters of 20% solution is 8(0.20) = 1.6 liters. cont’d

18. 18 Example 4 – Mixing Acid The amount of acid in 12 liters of 30% solution is 12(0.30) = 3.6 liters. Since 2 + 1.6 = 3.6, the results check. cont’d

19. 19 Solve a dry mixture application using a linear equation in one variable 3.

20. 20 Example – Mixing nuts Fancy cashews are not selling at $9 per pound, because they are too expensive. However, filberts are selling well at $6 per pound. How many pounds of filberts should be combined with 50 pounds of cashews to obtain a mixture that can be sold at $7 per pound? 1.What am I asked to find? Number of pounds of filberts in the mixture: x Total number of pounds of mixture: 50 + x

21. 21 Example – Mixing nuts 2.Form an equation Dry mixture problems are based on the formula v = pn, where v is the value of the mixture, p is the price per pound, and n is the number of pounds. Value of filberts: $6x Value of cashews: $9(50) = $45 Value of mixture: $7(50 + x) cont’d

22. 22 Example – Mixing nuts The value of the filberts (in dollars) 6x plus the value of the cashews (in dollars) 450, is equal to the value of the mixture (in dollars) 7(50 + x). We can organize this information in a table or a diagram, as shown in Figure 2-15. (a) (b) Figure 2-15 cont’d

23. 23 Example – Mixing nuts We can form the equation: 4.Solve the equation. 6x + 9(50) = 7(50 + x) 6x + 450 = 350 + 7x 100 = x (lb of filberts) cont’d

24. 24 Example – Mixing nuts 4.Check the result Filberts:100 pounds @ $6/lb: $600 Cashews: 50 pounds @ $9/lb: $450 Total: $1050 Mixture: 150 pounds @ $7/lb: $1,050. cont’d