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

2. The perimeter of a rectangle is 46 2L + 2W = 46
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. Which method should be used to solve this system of equations?
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. The length is 13 ft and the width is 10 ft.
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 W = 46
4W + 6 = 46
4w (-6) = 46 + (-6)
4W = 40
W = 10
The length is 13 ft and the width is 10 ft.
Back substitution
L = W + 3
L =
L = 13

5. The length is 5 ft and the width is 20 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
The length is 5 ft and the width is 20 ft.
Back substitution
W = 4L
W = 4(5)
W = 20