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Chapter 4 Systems of Linear Equations and Inequalities

Chapter 4 Systems of Linear Equations and Inequalities. Section 5 Solving Mixture Problems Using Systems of Linear Equations. Section 4.5 Objectives. 1 Draw Up a Plan for Modeling Mixture Problems 2 Set Up a Plan and Solve Money Problems Using the Mixture Model

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Chapter 4 Systems of Linear Equations and Inequalities

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  1. Chapter 4 Systems of Linear Equations and Inequalities Section 5 Solving Mixture Problems Using Systems of Linear Equations

  2. Section 4.5 Objectives 1 Draw Up a Plan for Modeling Mixture Problems 2 Set Up a Plan and Solve Money Problems Using the Mixture Model 3 Set Up and Solve Dry Mixture and Percent Mixture Problems

  3. Mixture Problems Problems that involve mixing two or more substances are called mixtureproblems. This type of problem can be set up using the model number of units of the same kind · rate = amount A table is helpful for organizing the given information.

  4. Mixture Problems Example: Matthew received an $8000 Christmas bonus from his employer. He invested part of the bonus in a savings account earning 3% simple interest per year and the rest in a certificate of deposit (CD) that earns 4.5% simple interest annually. At the end of the year, he will receive $330 interest on his investments. How much money did Matthew invest in each account? Step 1: Identify We want to know how much money was put into the saving account at 3% interest and how much was put into the CD earning 4.5% interest. Step 2: Name Let x = the amount of money deposited into the savings account. Let y = the amount of money deposited into the CD. Continued.

  5. Mixture Problems Example continued: Step 3: Translate Summarize the information in a table. Total amount invested x + y = $8000 The interest earned at 3% is 0.03x. The interest earned at 4.5% is 0.045y. Total interest earned 0.03x + 0.045y = $330 Continued.

  6. Mixture Problems Example continued: The system of equations is (1) (2) x + y = 8000 0.03x + 0.045y = 330 Step 4: Solve The elimination method will work easily. – 0.03( x + y = 8000) Multiply equation (1) by – 0.03. Simplify. – 0.03x – 0.03y = – 240 Add equation (1) and (2). 0.03x + 0.045y = 330 Simplify. 0.015y = 90 Divide both sides by 0.015. y = 6000 Continued.

  7. Mixture Problems Example continued: y = 6000 x + 6000 = 8000 Substitute 6000 into equation (1). x = 2000 Subtract 6000 from both sides. Step 5: Check x + y = 8000 0.03x + 0.045y = 330 2000 + 6000 = 8000 0.03(2000) + 0.045(6000)= 330 8000 = 8000 60 + 270 = 330  330 = 330  Step 6: Answer $2000 was invested in the savings account earning 3% simple interest per year and $6000 was invested in the CD earning 4.5% simple interest annually.

  8. Mixture Problems Example: A nut company wants to produce 20 pounds of nuts worth $4.50 per pound. To obtain this, they want to mix nuts worth $5.00 per pound with nuts worth $3.00 per pound. How many pounds of each type should be used to make the desired 20 pounds worth $4.50 per pound? Step 1: Identify We want to know how many pounds of each type of nuts to use to obtain 20 pounds of nuts worth $4.50 per pound. Step 2: Name Let x = the number of pounds of nuts worth $5.00 per pound. Let y = the number of pounds of nuts worth $3.00 per pound. Continued.

  9. Mixture Problems Example continued: Step 3: Translate Summarize the information in a table. Total pounds of nuts x + y = 20 Total value of the 20 pounds 5x + 3y = 4.5(20) Continued.

  10. Mixture Problems Example continued: The system of equations is x + y = 20 (1) 5x + 3y = 90 (2) Step 4: Solve x = 20 – y Solve equation (1) for x. Substitute 20 – y into (2). 5(20 – y)+ 3y = 90 Distribute. 100 – 5y + 3y = 90 Simplify. – 2y = – 10 Simplify. y = 5 Substitute y = 5 into (1). x + 5 = 20 Simplify. x = 15 Continued.

  11. Mixture Problems Example continued: x = 15 y = 5 Step 5: Check x + y = 20 5x + 3y = 90 15 + 5 = 20 5(15) + 3(5) = 90 20 = 20 75 + 15 = 90  90 = 90  Step 6: Answer 15 pounds of the $5.00 per pound nuts and 5 pounds of the $3.00 per pound nuts should be used to obtain 20 pounds of nuts worth $4.50 per pound.

  12. Mixture Problems Example: You work in the chemistry stockroom and your instructor has asked you to prepare 4 liters of 15% hydrochloric acid (HCl). Looking through the supply room you see that there is a bottle of 12% HCl and another of 20% HCl. How much of each should you mix so that your instructor has the required solution? Step 1: Identify We want to know the number of liters of 12% HCl that must be mixed with a 20% HCl solution to prepare a 15% HCl solution. We need 4 liters. Step 2: Name Let x = the number of liters of 12% HCl solution. Let y = the number of liters of 20% HCl solution. Continued.

  13. Mixture Problems Example continued: Step 3: Translate Summarize the information in a table. Total Liters x + y = 4 Pure HCl 0.12x + 0.20y = (0.15)(4) Continued.

  14. Mixture Problems Example continued: The system of equations is x + y = 4 (1) 0.12x + 0.20y = 0.6 (2) Step 4: Solve y = 4 – x Solve equation (1) for y. Substitute 20 – y into (2). 0.12x + 0.20(4 – x)= 0.6 Distribute. 0.12x + 0.8 – 0.20x = 0.6 Simplify. – 0.08x = –0.2 Simplify. x = 2.5 Substitute x = 2.5 into (1). 2.5 + y = 20 Simplify. y = 1.5 Continued.

  15. Mixture Problems Example continued: x = 2.5 y = 1.5 Step 5: Check x + y = 20 0.12x + 0.20y = 0.6 2.5 + 1.5 = 4 0.12(2.5) + 0.20(1.5) = 90 4 = 4 0.3 + 0.3 = 0.6  0.6 = 0.6  Step 6: Answer 2.5 liters of 12% HCl must be mixed with 1.5 liters of 20% HCl to form the required 4 liters of 15% HCl.

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