Question

Find the real solution of the system of equations: \[ x^4 + y^4 - x^2 y^2 = 13,\quad x^2 - y^2 + 2xy = 1 \] Satisfying condition: \( xy \geq 0 \)

• A. \( (x = 1, y = -2); (x = -1, y = 2) \)
• B. \( (x = 2, y = 1); (x = -2, y = -1) \)
• C. \( (x = 1, y = 2); (x = -1, y = -2) \)
• D. \( (x = 1, y = -2); (x = -1, y = -2) \)

Answer 1

The Correct Option is C

Try values: For \(x = 1, y = 2\): LHS1: \(1 + 16 - 4 = 13\), LHS2: \(1 - 4 + 4 = 1\), OK. Also \(xy = 2>0\) Similarly, \(x = -1, y = -2\) ⇒ works too.