SOLUTION: A carpenter used 26 ft of molding in three pieces to trim a garage door. If the long piece was 5 ft longer than twice the length of each shorter piece, what is the length of the lo

Question 27687: A carpenter used 26 ft of molding in three pieces to trim a garage door. If the long piece was 5 ft longer than twice the length of each shorter piece, what is the length of the longest piece?

Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
A carpenter used 26 ft of molding in three pieces to trim a garage door. If the long piece was 5 ft longer than twice the length of each shorter piece, what is the length of the longest piece?
The problem is not very well worded. Any way we assume that the shorter pieces are of the same length say S ft
Let the three pieces be L,S and S ft respectively.
The long piece is 5 ft longer than twice the length of each shorter piece
That is L is 5 ft longer than 2S
That is L = 2S +5
L-2S = 5 -----(1)
And L+S+S = 26 by data
That is L+2S = 26 ----(2)
(1) + (2) implies
(L-2S) +(L+2S) = 5 + 26 (adding the left with left and the right with right)
(L+L) = 31
(using additive commutativity and associativity
and the additive inverse law: (-2S+2S) =0)
Therefore 2L = 31 which means L = 31/2 = 15.5 ft
(2) implies L+2S = 26
Putting L = 15.5 in this
15.5 + 2S = 26
2S = 26.0-15.5
2S = 10.5 = 10.50
2S = 10.50
S = 10.50/2 = 5.25 feet
Therefore the three pieces are 15.5 ft, 5.25 ft and 5.25 ft and the longest piece is 15.5 ft
Verification:
15.5 + 5.25 + 5.25 = 26 which is correct.

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