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5-2B Word Problems - Solving Linear System by Substitution

5-2B Word Problems - Solving Linear System by Substitution. Algebra 1 Glencoe McGraw-Hill Linda Stamper.

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5-2B Word Problems - Solving Linear System by Substitution

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  1. 5-2B Word Problems - Solving Linear System by Substitution Algebra 1 Glencoe McGraw-Hill Linda Stamper

  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. Solving Word Problems Using A Linear System 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.

  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. 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? NumberLabels. Let b = # of book boxes Let c = # of clothing boxes weight of book boxes weight of the clothing boxes Let 60b = WeightLabels Let 15c = Verbal Model (to represent the number of boxes) # of clothing boxes + # of book boxes = Total # Equation b = c + 8 Verbal Model (to represent the weight of boxes) Total Weight Weight of clothing boxes Weight of book boxes = + 255 Equation 60b 15c = + Solve the linear system.

  6. Solve the linear system. 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. Jenny has 3 book boxes and 5 clothing boxes . Word problems require word answers. Sentence.

  7. 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? 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.

  8. Example 1In 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 a = # of adults Let c = # of children Number Labels. value of children’s tickets value of adult tickets Let 6a = ValueLabels Let 4c =

  9. Example 1In 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 a = # of adults Let c = # of children Number Labels. value of children’s tickets value of adult tickets Let 6a = ValueLabels Let 4c = Verbal Model (to represent the number of people) # of children = # of adults Total # of people + Equation a + c 321 = Verbal Model (to represent the value of tickets/admission) Value of children’s tickets Total Value Value of adult tickets + = Equation 4c + = 6a 1590 Solve the linear system.

  10. Solve the linear system. 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?

  11. Example 2An 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? Number Labels Let a = # of Stock A Let b = # of Stock B ValueLabels Let 75b = value of Stock B Let 50a = value of Stock A

  12. Example 2An 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? Number Labels Let a = # of Stock A Let b = # of Stock B ValueLabels Let 75b = value of Stock B Let 50a = value of Stock A Verbal Model (to represent the number of stock) Total # of stock + # of Stock B = # of Stock A 225 a = Equation + b Verbal Model (to represent the value of stock) Value of Stock A + Value of Stock B = Total Value of stock + 50a Equation 13,750 75b = Solve the linear system.

  13. Solve the linear system. The investor purchased 125 shares of Stock A and 100 shares of Stock B.

  14. Example 3The 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 letl =length width width length Formula Length = 2 widths + 1 2 lengths + 2 widths = perimeter The width is 18 m and the length is 37 m.

  15. Practice Problems • 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?

  16. Practice Problems • 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.

  17. Homework 5-A4 Skills Practice Wkb. Page 32 #7-20

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