Question 1098118: The length and width of a rectangle are represented by x+9 and 9-4x. If x must be an integer, what are the possible values for the area of the Rectangle?
Found 3 solutions by Alan3354, ikleyn, MathTherapy:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! The length and width of a rectangle are represented by x+9 and 9-4x. If x must be an integer, what are the possible values for the area of the Rectangle?
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9-4x has to be positive.
--> x < 3
==> x = 0, 1 or 2.
etc
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
Possible values for x are
x = 2, 1, 0, -1, -2, -3, -4, -5, -6, -7 and -8.
The table of values
x length width Area
2 11 1 11
1 10 5 50
0 9 9 81
-1 8 13 104
-2 7 17 119
-3 6 21 126
-4 5 25 125
-5 4 29 116
-6 3 33 99
-7 2 37 74
-8 1 41 41
Looking into solutions of many other tutors, I conclude that today is BAD day for many of them.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
The length and width of a rectangle are represented by x+9 and 9-4x. If x must be an integer, what are the possible values for the area of the Rectangle?
Largest value needed to keep 9 - 4x POSITIVE is 2, as the width would then be 1 (smallest POSITIVE integer).
Smallest value needed to keep x + 9 POSITIVE is - 8, as the length would then be 1 (smallest POSITIVE integer).
In addition, we also have all other integers BETWEEN these 2 values, or: 
That's it!
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