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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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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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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

Note that the expression y + xy can be factored as y(1 + x). Let’s simplify the given equation:

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

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

The two terms on the left side of the equation have (1 + x) in common, so we can factor it out:

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

Since x and y are positive integers, so are (1 + x) and (1 + y). We should investigate the different ways of writing 15 as a product of positive integers. Note that 15 can be expressed as a product of positive integers in two ways: 1 x 15 (scenario one) or 3 x 5 (scenario two).

For scenario one, we have either 1 + x = 1 and 1 + y = 15 OR 1 + x = 15 and 1 + y = 1. Note that in either case, one of x or y equals 0; since x and y are given as positive, this scenario is ruled out.

For scenario two, we have either 1 + x = 3 and 1 + y = 5, which means x = 2 and y = 4 OR 1 + x = 5 and 1 + y = 3, meaning x = 4 and y = 2. In either case, x + y = 6.

Answer: C
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