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

The length is 3 ft more than the width. L = W + 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

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

Back substitution L = W + 3 L = 10 + 3 L = 13

The length is 13 ft and the width is 10 ft.

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

Back substitution W = 4L W = 4(5) W = 20

The length is 5 ft and the width is 20 ft.