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1+x+y+xy

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by DeepakR » Sun Apr 19, 2009 3:35 am
I just picked the answer choices and concluded the solution.

x+y=3 has the following combinations of x and y= (1,2) or (2,1) both will not satisfy the given equation. But if you have x+y=6 the following are the combinations (4,2) or (5,1) etc of which (4,2) will satisfy the given equation 1+ 4+ 2+ (4*2)=15. Hence C.) 6

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by DanaJ » Sun Apr 19, 2009 4:08 am
I'd go for a more mathematical solution.

1 + x + y + xy = (x + 1)(y + 1) = 15.
Now, there are only two situations for x + 1 and y + 1, since they both have to be positive:
a. 1*15 = 15. In this case, you get that x + 1 is 1 or 15 and y + 1 is 15 or 1. But, since x and y are both positive, this is not a viable solution: if x + 1 = 1, then x = 0, meaning that x is not positive. Same goes for the case when you have y + 1 = 15.

b. 3*5 = 15. In this case, you get x + 1 = 3 or 5 and y + 1 = 5 or 3. In any case, x will be 2 or 4 and y will be 4 or 2. This means that whatever the situation, the sum x + y will always be 6.
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by lilu » Sun Apr 19, 2009 9:53 am
I also picked numbers.
x+y+xy=14
So, there not that many numbers we need to look at in order to figure out the solution since the sum of these 3 integers is pretty small. Also, both numbers have to be even.
So, if we try x=2 and y=2, the number we get is too small.
So, then we try x=2 and y=4 (the fact that we need to find the sum also makes the task much easier-->we don't need to try y=4 and x=2) and bingo, they give the right combination.
So, their sum is 6
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by dumb.doofus » Sun Apr 19, 2009 10:29 am
The equation 1 + x + y + xy = 15 can be written as

1 + x + y(1 + x) = 15

--> (1+x)(1+y) = 15


So now we have two solutions, (3,5) and (1,15)
Now a positive integer is any integer that is greater than 0, so the only solution that we are left with is (3,5), which implies that
x+y = 6 i.e. C
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