1. 4.2B Word Problems - Solving Linear System by Substitution

2. Previously when solving word problems you were restricted to one variable. Today you will solve word problems using more than one variable. If you use more than one variable, you will need to write a system of linear equations.

3. 1) Write two sets of labels, if necessary (one set for number, one set for value, weight etc.) 2) Write two verbal models. (Translate from sentences) 3) Write two algebraic models (equations). 4) Solve the system of equations. 5) Write a sentence and check your solution in the word problem. Solving Word Problems Using A Linear System

4. Jenny has 8 moving boxes that she can use to pack for college. Each box can hold 15 pounds of clothing or 60 pounds of books. If Jenny is moving 255 pounds how many boxes of each type are there?

5. Let c = # of clothing boxes Jenny has 8 moving boxes that she can use to pack for college. Each box can hold 15 pounds of clothing or 60 pounds of books. If Jenny is moving 255 pounds how many boxes of each type are there? Let 15c = Let b = # of book boxes Number Labels. Weight Labels Let 60b = weight of the clothing boxes weight of book boxes c+ b = 8 Verbal Model (to represent the number of boxes) # of clothing boxes + # of book boxes = Total # Equation 15c15c + 60b = 255 Verbal Model (to represent the weight of boxes) Weight of clothing boxes Equation Weight of book boxes Total Weight + = Solve the linear system.

6. Choose one equation and isolate one of the variables. Substitute the expression into the other equation and solve. Substitute the solved value into one of the original equations and solve. Sentence. Jenny has 3 book boxes and 5 clothing boxes. Could the answer be 2 book boxes and 6 clothing boxes? Check the solution in the word problem. Word problems require word answers.

7. Let a = # of adults Example 1 In one day the museum collected $1590 from 321 people. The price of admission is $6 for an adult and $4 for a child. How many adults and how many children were admitted to the museum? Let 6a = Let c = # of children Number Labels. Value Labels Let 4c = value of adult tickets value of children ’ s tickets

8. Let a = # of adults Example 1 In one day the museum collected $1590 from 321 people. The price of admission is $6 for an adult and $4 for a child. How many adults and how many children were admitted to the museum? Let 6a = Let c = # of children Number Labels. Value Labels Let 4c = value of adult tickets value of children ’ s tickets a + c = 321 Verbal Model (to represent the number of people) # of adults Equation # of children Total # of people + = 6a + 4c4c = 1590 Verbal Model (to represent the value of tickets/admission) Value of adult tickets Equation Value of children ’ s tickets Total Value + = Solve the linear system.

9. The museum admitted 153 adults and 168 children. Are there other values that will total 321? How do you know if you have the correct combination?

10. Let a = # of Stock A Example 2 An investor bought 225 shares of stock. Stock A was purchased at $50 per share and Stock B at $75 per share. If $13,750 worth of stock was purchased, how many shares of each kind did the investor buy? Let 50a = value of Stock A Let b = # of Stock B Number Labels Value Labels Let 75b = value of Stock B

11. Let a = # of Stock A Example 2 An investor bought 225 shares of stock. Stock A was purchased at $50 per share and Stock B at $75 per share. If $13,750 worth of stock was purchased, how many shares of each kind did the investor buy? Let 50a = value of Stock A Let b = # of Stock B Number Labels Value Labels Let 75b = value of Stock B a + b = 225 Verbal Model (to represent the number of stock) # of Stock A Equation # of Stock B Total # of stock + = 50a + 75b = 13,750 Verbal Model (to represent the value of stock) Value of Stock A + Value of Stock B = Total Value of stock Equation Solve the linear system.

12. The investor purchased 125 shares of Stock A and 100 shares of Stock B.

13. Example 3 The length of a rectangle is 1 m more than twice its width. If the perimeter is 110 m, find the dimensions. Length let w = width let l = length Formula length width 2 lengths + 2 widths = perimeter The width is 18 m and the length is 37 m. = 2 widths + 1 =

14. Practice Problems 1.A sightseeing boat charges $5 for children and $8 for adults. On its first trip of the day, it collected $439 for 71 paying passengers. How many children and how many adults were there? 2.The length of a rectangle is 12 inches more than twice its width. If the perimeter is 90 inches, what are the dimensions of the rectangle?

15. Practice Problems 1.A sightseeing boat charges $5 for children and $8 for adults. On its first trip of the day, it collected $439 for 71 paying passengers. How many children and how many adults were there? There were 43 children and 28 adults on the sightseeing trip. 2.The length of a rectangle is 12 inches more than twice its width. If the perimeter is 90 inches, what are the dimensions of the rectangle? The width is 11 inches and the length is 34 inches.

16. 7-A3 Pages 399-401 # 24–33, 48–52.