Verified answer

x - y = 6

x = y + 6

substitute y + 6 for x in the other equation:

(y + 6)^2 + y^2 = 50

y^2 + 12y + 36 + y^2 = 50

2y^2 + 12y + 36 = 50

2y^2 + 12y - 14 = 0

y^2 + 6y - 7 = 0

(y + 7)(y - 1) = 0

y + 7 = 0 or y - 1 = 0

y = -7 or y = 1

Since y is positive, discard y = -7, leaving just

y = 1

x = 1 + 6 = 7

xy = (1)(7) = 7

If x - y = 6, then

y = x - 6

Substitute this for y into

x^2 + y^2 = 50

So that

x^2 + (x - 6)^2 = 50

x^2 + (x^2 - 12x + 36) = 50

On subtracting 50 from both sides:

2x^2 - 12x - 14 = 0

Divide both sides by 2

x^2 - 6x - 7 = 0

Factorising

(x+1)(x-7) =0

So x has two solutions

x = -1 and x = 7

For x = -1

-1 - y = 6

=> y = -7

and xy = (-1)(-7) = 7

But we are given that both x and y are positive integers. So

x = -1 and y = -7 are not solutions

For x = 7

7 - y = 6

=> y = 1

and xy = 7*1 = 7

50 - 2xy = 36

- 2xy = - 14

xy = 7