6. If \( x \) and \( y \) are positive integers, \( x^{2}-y^{2}=\frac{8}{9} \), and \( x^{2}+2 x y+y^{2}=\frac{16}{9} \), what is the value of \( x-y \) ?

To solve the equations given in the problem, we start with the two equations: 1. \( x^2 - y^2 = \frac{8}{9} \) 2. \( x^2 + 2xy + y^2 = \frac{16}{9} \) We can rewrite the first equation using the difference of squares: \[ (x - y)(x + y) = \frac{8}{9} \] Next, we can rewrite the second equation. The expression \( x^2 + 2xy + y^2 \) can be factored as: \[ (x + y)^2 = \frac{16}{9} \] Taking the square root of both sides gives: \[ x + y = \frac{4}{3} \quad \text{or} \quad x + y = -\frac{4}{3} \] Since \( x \) and \( y \) are positive integers, we discard the negative solution and take: \[ x + y = \frac{4}{3} \] Now we have two equations: 1. \( (x - y)(x + y) = \frac{8}{9} \) 2. \( x + y = \frac{4}{3} \) Substituting \( x + y = \frac{4}{3} \) into the first equation gives: \[ (x - y) \left(\frac{4}{3}\right) = \frac{8}{9} \] To isolate \( x - y \), we multiply both sides by \( \frac{3}{4} \): \[ x - y = \frac{8}{9} \cdot \frac{3}{4} = \frac{8 \cdot 3}{9 \cdot 4} = \frac{24}{36} = \frac{2}{3} \] Now we have: \[ x - y = \frac{2}{3} \] However, since \( x \) and \( y \) are positive integers, \( x - y \) must also be an integer. The value \( \frac{2}{3} \) cannot be an integer, which indicates that we need to re-evaluate our approach. Let's go back to the equations and consider the integer nature of \( x \) and \( y \). We can express \( x \) and \( y \) in terms of \( a \) and \( b \) where \( a = x - y \) and \( b = x + y \): From \( b = \frac{4}{3} \), we can express \( x \) and \( y \): \[ x = \frac{a + b}{2} = \frac{a + \frac{4}{3}}{2} \] \[ y = \frac{b - a}{2} = \frac{\frac{4}{3} - a}{2} \] Since \( x \) and \( y \) must be integers, both \( a + \frac{4}{3} \) and \( \frac{4}{3} - a \) must be even integers. This leads us to conclude that \( a \) must be a multiple of \( \frac{2}{3} \), which is not possible for integers. Thus, we need to find integer solutions for \( x \) and \( y \) that satisfy both equations. Let's try some integer values for \( x \) and \( y \): 1. If \( x = 2 \) and \( y = 1 \): - \( x^2 - y^2 = 4 - 1 = 3 \) (not equal to \( \frac{8}{9} \)) - \( x^2 + 2xy + y^2 = 4 + 4 + 1 = 9 \) (not equal to \( \frac{16}{9} \)) 2. If \( x = 3 \) and \( y = 1 \): - \( x^2 - y^2 = 9 - 1 = 8 \) (not equal to \( \frac{8}{9} \)) - \( x^2 + 2xy + y^2 = 9 + 6 + 1 = 16 \) (not equal to \( \frac{16}{9} \)) Continuing this process, we find that no integer solutions satisfy both equations. Thus, we conclude that the problem as stated does not yield integer solutions for \( x \) and \( y \) under the given conditions. Therefore, the value of \( x - y \) cannot be determined as a positive integer. The final answer is: \[ \