Dari persamaan (x² + 1)(y² + 1) + 9 = 6(x + y), nilai dari x² + y² = 7.
(dengan x, y ∈ ℝ)
 

Penjelasan dengan langkah-langkah:

(x² + 1)(y² + 1) + 9 = 6(x + y)
⇒ x²y² + x² + y² + 1 + 9 = 6(x + y)
⇒ (xy)² + (x + y)² – 2xy + 1 + 9 = 6(x + y)
⇒ (xy)² – 2xy + 1 + (x + y)² + 9 = 6(x + y)
⇒ (xy – 1)² + (x + y)² + 9 = 6(x + y)
⇒ (xy – 1)² + (x + y)² – 2(x + y) + 1 = 4(x + y) – 8
⇒ (xy – 1)² + [(x + y) – 1]² = 4(x + y) – 8
⇒ (xy – 1)² + [(x + y) – 1]² = 4[(x + y) – 1] – 4
⇒ (xy – 1)² + [(x + y) – 1]² – 4[(x + y) – 1] + 4 = 0
⇒ (xy – 1)² + [((x + y) – 1) – 2]² = 0
⇒ (xy – 1)² + (x + y – 3)² = 0
⇒ (xy – 1)² = –(x + y – 3)²

Jika x, y ∈ ℝ, karena (xy – 1)² ≥ 0, maka –(x + y – 3)² = 0, sehingga
(xy – 1)² = 0  dan  (x + y – 3)² = 0.

• (xy – 1)² = 0
  ⇒ xy – 1 = 0
  ⇒ xy = 1

• (x + y – 3)² = 0
  ⇒ x + y – 3 = 0
  ⇒ x + y = 3

Oleh karena itu:
x² + y² = (x + y)² – 2xy
⇒ x² + y² = 3² – 2·1
⇒ x² + y² = 7