SOLUTION: The length of a wall is 3m more than its witdh.
If the area of the wall is less than 18m squared,
What could be its lenght?
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-> SOLUTION: The length of a wall is 3m more than its witdh.
If the area of the wall is less than 18m squared,
What could be its lenght?
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Question 1186358: The length of a wall is 3m more than its witdh.
If the area of the wall is less than 18m squared,
What could be its lenght? Answer by greenestamps(13200) (Show Source):
The problem is far more easily solved informally than with formal algebra.
6 times 3 is 18, and 3 is 3 less than 6; so the maximum length is 6.
The minimum length is whatever value keeps the width positive; since the width is 3 less than the length, the minimum length is 3.
ANSWER: The length can be any x in the interval (3,6)
Note the length can't be 6, because the area has to be LESS THAN 18 square meters; and it can't be 3, because the width would be 0 and there would be no wall.
If formal algebra is required....
x = length
x-3 = width
The area (length times width) has to be less than 18:
Algebraically, the solution is x between -3 and 6....
But in the actual problem, since the width has to be positive (x-3>0 --> x>3), the real solution is that x is between 3 and 6.
