Java X^2-y^2=16& √x-√y=√2. What is the value of x+y?

 To tackle this arrangement of conditions, we should signify √x as an and √y as b. Then, at that point, we have the accompanying framework:

1. \(a^2 - b^2 = 16\)

2. \(a - b = \sqrt{2}\)

Presently, we should tackle for an and b. Adding the two conditions, we get \(a^2 - b^2 + a - b = 16 + \sqrt{2}\). Calculating the left side, we have \((a + 1)(a - b) = 16 + \sqrt{2}\).

Since \(a - b = \sqrt{2}\), we can substitute this into the situation: \((a + 1)\sqrt{2} = 16 + \sqrt{2}\). Tackling for a, we find \(a = 7 + 4\sqrt{2}\).

Since we have \(a\), we can find \(b\) by deducting \(\sqrt{2}\): \(b = a - \sqrt{2} = 6 + 4\sqrt{2}\).

At long last, \(x = a^2\) and \(y = b^2\), so \(x + y = a^2 + b^2\).

Ascertaining \(x + y\), we get \(x + y = (7 + 4\sqrt{2})^2 + (6 + 4\sqrt{2})^2\). The worth of \(x + y\) is around \(170.97\).