if (x²+y²) varies dircectly as (x²-y²)

The answer from the above respondent only tells you (x + y) varies directly as (x² + y²)/(x - y). If we want to prove (x+y) varies directly as (x-y), we have to remove all variables except (x+y) and (x-y).

If (x²+y²) varies directly as (x²-y²), we have
(x² + y²) = k(x² - y²), where k is a non-zero constant.

y² + ky² = kx² - x²
y² (1 + k) = x² (k - 1)
Thus x² = (1+k)/(k-1) X y²
Note that (1+k)/(k-1) is in fact another constant, so let c² = (1+k)/(k-1).

We have x² = c²y²
Taking square roots on both sides, x = cy. In other words, x varies directly as y.

Consider (x+y) / (x-y)

(x+y) / (x-y)
= (cy + y) / (cy - y)
= [y(c+1)] / [y(c-1)]
= (c+1) / (c-1)

Note that c is a constant, (c+1) / (c-1) is also a constant.
That is, (x+y) / (x-y) = constant.

Therefore (x+y) varies directly as (x-y).

Because (x² + y²) varies dircectly as (x² - y²)

So, let x² + y² = k(x² - y²) , where k is a constant.


i.e. x² + y² = k(x + y)(x - y)

x + y = (x² + y²)/k(x - y)

x + y = 1/k X (x² + y²)/(x - y)


So, if (x² + y²) varies dircectly as (x² - y²)

(x + y) varies inversely as (x - y).