Question 997418: The Perimeter of a rectangle is 54m. If the width is doubled and the length were increased by 15m, the perimeter would be 98m. What are the length and width of the rectangle?
it says to solve via substitution.
I keep getting 14.5 for x, yet on my answer sheet its wrong. I must be doing something incorrect.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! The Perimeter of a rectangle is 54m.
2L + 2W = 54
simplify, divide by 2
L + W = 27
L = (27-W); use this form for substitution
:
If the width is doubled and the length were increased by 15m, the perimeter would be 98m.
2(L+15) + 2(2W) = 98
simplify, divide by 2
(L+15) + 2W = 49
L + 15 + 2W = 49
Substitute (27-W) for L
(27-W) + 15 + 2W = 49
W + 42 = 49
W = 49 - 42
W = 7 m is the width
then
L = 27 - 7
L = 20 m is the length
:
:
Check this by finding the perimeter with these value
2(20) + 2(7) =
40 + 14 = 54
:
What are the length and width of the rectangle? 20 by 7
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