Let x = greater number
y = smaller number
(1) \(x = y + 6\)
(1) \(y = x-6\)
(2) \(x^{2} + y^{2} = 90\)
We'll substitute y in (1) to (2)
(2) \(x^{2} + (x-6)^{2} = 90\)
\(x^{2} + (x^{2} - 12x + 36) = 90\)
\(x^2 + x^2 -12x + 36 - 90 = 0\)
\(2x^2 - 12x -54 = 0\)
\(2(x^2 - 6x - 27) = 0\)
\(2(x - 9)(x + 3) = 0\)
x - 9 = 0 or x + 3 = 0
x = 9 x = -3
and
y = x - 6 y = x - 6
y = 9 - 6 y = -3 - 6
y = 3 y = -9
Therefore, the two numbers can be 9 and 3 or -3 and -9.
