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Total-Value Problems

Total-Value Problems. Example Maya sells concessions at a local sporting event. In one hour, she sells 72 drinks. The drink sizes are small, which sells for $2 each, and large, which sells for $3 each. If her total sales revenue was $190, how many of each size did she sell?

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Total-Value Problems

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  1. Total-Value Problems • Example Maya sells concessions at a local sporting event. In one hour, she sells 72 drinks. The drink sizes are small, which sells for $2 each, and large, which sells for $3 each. If her total sales revenue was $190, how many of each size did she sell? • 1. Introduction. We let • s = the number of small drinks l = the number of large drinks. 2. Body. Organize the information in a table. This can be helpful with translating. l + s = 72 3l + 2s = 190 We are to solve the system of equations

  2. continued Next, we replace l in equation (2) with l = 72  s: We find l by substituting 26 for s in equation (3): l = 72  s = 72  26 = 46. So, the solution is (46, 26).

  3. Menu 1 EDIT Matrix Solution of Systems by Calculator Write the system of equations as an augmented matrix Input data

  4. Matrix Solution of Systems by Calculator Menu 1 Menu B:rref( So the solution to Menu 1

  5. continued—Graphing calculator approach First we create an augmented matrix from the system. l+ s= 72 3l+2s= 190 Next we use our calculator to put the matrix into reduced row echelon form. The point of intersection is (46, 26). We have l= 26 and s= 46. 3.Conclusion. Maya sold 46 large drinks and 26 small drinks. Check. If Maya sold 46 large drinks and 26 small drinks, she would have sold 72 drinks, for a total of 46($3) + 26($2) = $138 + $52 = $190.

  6. c + r = 37 26c + 34r = 1138 Example A cookware consultant sells two sizes of pizza stones. The circular stone sells for $26 and the rectangular one sells for $34. In one month she sold 37 stones. If she made a total of $1138 from the sale of the pizza stones, how many of each size did she sell? 1. Introduction. We let, c = number of circular stone sold r = number of rectangular stone sold Presenting the information in a table can be helpful. 2. Body. We have translated to a system of equations: c + r = 37 26c + 34r = 1138

  7. continued—Graphing Calculator Solution Solve. The system can be solved by using an augmented matrix Next we use our calculator to put the matrix into reduced row echelon form. The point of intersection is (15, 22). We have c = 15 and r = 22. 3.Conclusion. The consultant sold 15 circular pizza stones and 22 rectangular pizza stones. Check. If c = 15 and r= 22, a total of 37 stones were sold. The amount paid was ($26) (15) + ($34) (22) = $1138. The numbers check.

  8. Mixture Problems i+ v = 60 9.95i + 11.25v = 630 Example A coffee shop is considering a new mixture of coffee beans. It will be created with Italian Roast beans costing $9.95 per pound and the Venezuelan Blend beans costing $11.25 per pound. The types will be mixed to form a 60-lb batch that sells for $10.50 per pound. How many pounds of each type of bean should go into the blend? 1. Introduction. Let i= the number of pounds of Italian roast v = the number of pounds of Venezuelan blend Present the information in a table.

  9. 2. Body. We have translated to a system of equations: continued i + v = 60 9.95i + 11.25v = 630

  10. continued—Graphing Calculator Solution Solve. The system can be solved by using an augmented matrix Next we use our calculator to put the matrix into reduced row echelon form. The point of intersection is (34.6, 25.4). We have i = 34.6 and v = 25.4 . 3. Conclusion. The new blend should be made from 34.6 pounds of Italian Roast beans and 25.4 pounds of Venezuelan Blend beans.

  11. 20 liters t liters f liters Example An employee at a small cleaning company wishes to mix a cleaner that is 30% acid and another cleaner that is 50% acid. How many liters of each should be mixed to get 20 L of a solution that is 35% acid? Solution • Introduction. Let • t = the number of liters of the 30% solution • f = the number of liters of the 50% solution. + = 30% acid 50% acid 35% acid Amount of Acid (liters) 0.30t 0.50f 0.35(20) = 7 2. Body. We have the following system of equations:

  12. continued Solve. The system can be solved by using an augmented matrix Next we use our calculator to put the matrix into reduced row echelon form. We obtain (15, 5), or t = 15 and f = 5. 3. Conclusion. The employee should mix 15 L of the 30% solution with 5 L of the 50% solution to get 20 L of a 35% solution.

  13. Motion Problems When a problem deals with distance, speed (rate), and time, recall the following.

  14. Example • Two cars leave Toledo at the same time traveling in opposite directions. One car travels at 60 mph and the other at 45 mph. In how many hours will they be 225 miles apart? • 1. Introduction. Make a drawing. 60 mph 45 mph = r d t Use a table. 225 miles 60t 60 t 45t 45 t

  15. continued 2. Body. Distance of fast car + Distance of slow car = 225 From the table we get: 60t + 45t = 225 Solve. 3. Conclusion. In 2 1/7 hours the cars will be 225 miles apart.

  16. Example Because of a tail wind, a jet is able to fly 20 mph faster than another jet that is flying into the wind. In the same time that it takes the first jet to travel 90 miles the second jet travels 80 miles. How fast is each jet traveling?

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