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What's the approach to solve following example. If x and y are positive integers and 1 + x + y + xy = 15, what is the value of x + y? A. 3 B. 5 C. 6 D. 8 E. 9

\(1+x+y+xy=15\) --> \((x+1)(y+1)=15=3*5\);

As \(x\) and \(y\) are positive integers then \(x+1=3\) and \(y+1=5\), or vise-versa (x+1=1 and y+=15 or vise-versa is not possible as in this case one of the unknowns becomes zero and we are told that both unknowns are positive).

So \(x+1=3\) and \(y+1=5\), or vise-versa --> no need to solve for \(x\) and \(y\), just add them --> \(x+1+y+1=3+5\) --> \(x+y=6\).

I went by answe choices. x+y+xy=14 (given) If x+y = 3 then x=1, y 2 = cannot satisfy 14. If x+y=5, x=4, y=1 cannot satisfy 14. x=2 and y=3 ..ruled out x+y=6 = I got the answer

I liked the equation approach as its clean. Although I could solve it in less than a minute
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Re: If x and y are positive integers and 1 + x + y + xy = 15, wh [#permalink]

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02 Jan 2015, 02:07

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Re: If x and y are positive integers and 1 + x + y + xy = 15, [#permalink]

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If x and y are positive integers and 1 + x + y + xy = 15, what is the value of x + y?

A. 3 B. 5 C. 6 D. 8 E. 9

Easiest way to solve this question seems substitution because the values can't be too big as the multiplication of those two Positive Integers x and y must be less than 15 as per the given expression 1 + x + y + xy = 15

Values of x and y have to be less than 3 and 5 as 3*5=15

Let's try with 2 and 3 1+2+3+2*3 = 12

i.e. One of the numbers must be a little bigger

Let's try with 2 and 4 1+2+4+2*4 = 15 BINGO!!!

i.e. x+y = 2+4 = 6

Answer: option C
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Re: If x and y are positive integers and 1 + x + y + xy = 15, wh [#permalink]

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