SOLUTION: The perimeter of a rectangle is 62 m. If the width were doubled and the length were increased by 21 m, the perimeter would be 120 m. What are the length and width of the rectangle?

Question 173612This question is from textbook Introductory Algebra
: The perimeter of a rectangle is 62 m. If the width were doubled and the length were increased by 21 m, the perimeter would be 120 m. What are the length and width of the rectangle? This question is from textbook Introductory Algebra

Found 2 solutions by ankor@dixie-net.com, josmiceli:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The perimeter of a rectangle is 62 m. If the width were doubled and the length
were increased by 21 m, the perimeter would be 120 m. What are the length and
width of the rectangle?
:
Write an equation for each rectangle, and use elimination to solve this:
2L + 2W = 62
Simplify, divide by 2:
L + W = 31
:
and "If the width were doubled and the length were increased by 21 m," P = 120
:
2(L+21) + 2(2W) = 120
2L + 42 + 4W = 120
2L + 4W = 120 - 42
2L + 4W = 78
Simplify, divide by 2
L + 2W = 39
:
Subtract the 1st equation from the 2nd equation
L + 2W = 39
L + W = 31
-----------
W = 8 m
:
Find L using L + W = 31
L = 31 - 8
L = 23 m
:
:
Check solution by finding the perimeter of each
2(23) + 2(8) = 62
and
2(44) + 2(16) = 120

Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = length
Let = width
Let = perimeter

Given is
(1)
The problem says if I replace
with , and
with , then

(2)
Subtract (1) from (2)

Plug this result back into (1)

The length is 23 m and the width is 8 m
check:

OK

OK

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