Good morning Amelia. The first step to solving this problem would be to draw a rectangle and label the width and length. In a rectangle, two sides are equal and the other two sides are equal. This means that in your rectangle, there should be 2 sides that are labeled as the width (the shorter sides) and two sides that are labeled as the length (the longer sides). The perimeter is defined as the sum of all of the sides. In the problem, we know that the perimeter is 40. Now, we will define our variables. We should:
let W represent the width
let L represent the length
Since we have the 2 shortest sides labeled as W in our rectangle and the 2 longest sides labeled as L in our rectangle, our total width will now become 2W and our total length will now become 2L.
The problem also provides us with one more element that we will need to solve the problem. It states that "the length is 4 less than two times the width". This mathematical phrase would be expressed as:
L=2W-4
We can now set up our algebraic equation and solve for W.
2W+2L=40 (because the sum of all of the sides is 40)
2W + 2(2W-4) = 40 (replaced L with 2W-4)
2W+4W-8 = 40 (distributed 2 to both terms in the parentheses)
6W - 8= 40 (combined like terms, which were 2W and 4W)
6W= 48 (added 8 to both sides of the eqution)
W = 8 (divided by 6 on both sides to get the value of W)
Now that we have the value of W, we can solve for L by replacing W with 8 in the 2W+2L=40 equation.
2(8)+2L=40
16+2L=40
2L=24 (subtracted 16 from both sides of the equation)
L = 12 (divided by 2 on both sides of the equation)
If you have any questions, or want to schedule a lesson with me, please do not hesitate to contact me through Wyzant.
