What is the value of y?
(1) x^2 - y^2 = 5
(2) x and y are each positive integers
OA C
Source: Veritas Prep
What is the value of y?
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Each statement is clearly insufficient alone. Using both, from Statement 1 we can factor since we have a difference of squares:
(x + y)(x - y) = 5
From Statement 2, x+y and x-y are both integers, and if x and y are each positive integers, x+y is greater than 1. Since x+y is also clearly a factor of the prime number 5 from the equation above, x+y must equal 5 (since that's the only factor of 5 greater than one), and x-y must therefore equal 1, so we have two simple linear equations we can solve (x = 3 and y =2). The answer is C.
(x + y)(x - y) = 5
From Statement 2, x+y and x-y are both integers, and if x and y are each positive integers, x+y is greater than 1. Since x+y is also clearly a factor of the prime number 5 from the equation above, x+y must equal 5 (since that's the only factor of 5 greater than one), and x-y must therefore equal 1, so we have two simple linear equations we can solve (x = 3 and y =2). The answer is C.
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