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[tex]\large \sf \underline{Problem:}[/tex]
The perimeter of a rectangle is twice the
sum of its length and its width. The
perimeter is 40 meters and its length is 2
meters more than twice its width.
What is its length?
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[tex]\large \sf \underline{Answer:}[/tex]
The answer is 12 m.
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[tex]
\large \sf \underline{Explanation:}[/tex]
The perimeter (P) of a rectangle is twice the sum of its length (I) and width (w):
P = 2(l + w) P = 34
=>> 2(1+w) = 34
1+ w = 34/2
⇒ I + w = 17
Length is 2 meters more than twice its width:
I = 2w + 2
Now we have the system of equations:
I + w = 17
I = 2w + 2
Use the substitution method and substitute I from the second equation into the first equation:
2w + 2 + w = 17
2w + w = 17 - 2
3w = 15
w = 15/3
w = 5
Now, substitute w into the equation:
I = 2w + 2
I =2* 5+2
I = 10 + 2
I = 12
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