Question 239375: Solve using the five-step problem-solving process.
Find the length of a rectangular lot with a perimeter of 108 meters if the length is 8 meters more than the width. (P = 2L + 2W)
thanks
Answer by nyc_function(2741)   (Show Source):
You can put this solution on YOUR website! length = x + 8
width = x
P = 108
108 = 2(x + 8) + 2(x)
Can you finish now?
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I got your reply. Here is the rest of my reply.
length = x + 8
width = x
P = 108
108 = 2(x + 8) + 2(x)
We have the tools needed to solve for x and find the answer.
Keep in mind that we are searching for the length, which is 8 more than the width denoted as x + 8.
108 = 2(x + 8) + 2(x)
108 = 2x + 16 + 2x
108 = 4x + 16
108 - 16 = 4x
92 = 4x
92/4 = x
23 = x
Is the anwer 23?
No, it is not the answer. The number 23 is the value of x.
Since x = the width, we now know that the width is 23 meters.
Now, the length is x + 8 meters.
Replacing x with 23, we get 23 + 8 or 31 meters.
The length is 31 meters for this rectangular field.
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