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Solving systems of equations with 2 variables

Learn how to solve word problems involving perimeter of rectangles with two variables using the Substitution or Elimination Method. Example scenarios provided include finding dimensions given perimeter and relationships between length and width.

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Solving systems of equations with 2 variables

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  1. Solving systems of equations with 2 variables Word problems (Perimeter)

  2. 6)The perimeter of a rectangle is 46 feet. The length is 3 ft more than the width. Find the length and width. The perimeter of a rectangle is 46 2L + 2W = 46 The length is 3 ft more than the width. L = W + 3

  3. 6)The perimeter of a rectangle is 46 feet. The length is 3 ft more than the width. Find the length and width. 2L + 2W = 46 L = W + 3 Which method should be used to solve this system of equations? a) Substitution Method b) Elimination (Addition) Method

  4. 6)The perimeter of a rectangle is 46 feet. The length is 3 ft more than the width. Find the length and width. 2L + 2W = 46 L = W + 3 2(W + 3) + 2W = 46 2W + 6 + 2W = 46 4W + 6 = 46 4w + 6 + (-6) = 46 + (-6) 4W = 40 W = 10 The length is 13 ft and the width is 10 ft. Back substitution L = W + 3 L = 10 + 3 L = 13

  5. 7)The perimeter of a rectangle is 50 feet. The width is 4 times the length. Find the length and width. 2L + 2W = 50 W = 4L 2L + 2(4L) = 50 2L + 8L = 50 10L = 50 L = 5 The length is 5 ft and the width is 20 ft. Back substitution W = 4L W = 4(5) W = 20

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