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Solving systems of equations with 2 variables

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

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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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