SOLUTION: 3) The length of a rectangle is 2 in. more than the width. If the length and width are each increased by 3 in., the perimeter of the new rectangle will be 4 in. less than 8 times

Question 640428: 3) The length of a rectangle is 2 in. more than the width. If the length and width are each increased by 3 in., the perimeter of the new rectangle will be 4 in. less than 8 times the width of the original rectangle. Find the original dimensions of the rectangle.
Please help, I have no idea how to even start this!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The length of a rectangle is 2 in. more than the width.
If the length and width are each increased by 3 in., the perimeter of the
new rectangle will be 4 in. less than 8 times the width of the original rectangle.
Find the original dimensions of the rectangle.
:
Just write equation for what it says, step-by-step
:
"The length of a rectangle is 2 in. more than the width."
L = (W+2)
:
"If the length and width are each increased by 3 in."
(L+3) and (W+3)
"the perimeter of the new rectangle will be 4 in. less than 8 times the
width of the original rectangle.
2(L+3) + 2(W+3) = 8W - 4
2L + 6 + 2W + 6 = 8W - 4
2L + 12 = 8W - 2W - 4
2L = 6W - 4 - 12
2L = 6W - 16
simplify, divide by 2
L = 3W - 8
We know L = (W+2), therefore
w + 2 = 3W - 8
2 + 8 = 3W - W
10 = 2W
W = 10/2
W = 5 in is the original width
then
5+2 = 7 in is the original length
:
:
Confirm this by using these values in the statement
"the perimeter of the new rectangle will be 4 in. less than 8 times the width of the original rectangle."
2(7+3)) + 2(5+3) = 8(5) - 4
20 + 16 = 40 - 4
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