Question

Given the equations \( x + y = 1 \), \( x^2 + y^2 = 2 \), \( x^5 + y^5 = A \). Let N be the number of solution pairs \( (x, y) \) to this system of equations. Then \( N \) is equal to:

• A. 9
• B. 3
• C. \( \frac{7}{2} \)
• D. \( \frac{19}{2} \)

Hint:

For systems involving powers of sums, utilize algebraic identities like the expansion of \( (x + y)^n \) to simplify the problem.

Answer 1

The Correct Option is D

Step 1: Analyze the system of equations. 
We are given the equations: \[ x + y = 1 \text{(1)} \] \[ x^2 + y^2 = 2 \text{(2)} \] By squaring equation (1), we get: \[ (x + y)^2 = 1^2 = 1. \] Expanding this: \[ x^2 + 2xy + y^2 = 1. \] Substitute \( x^2 + y^2 = 2 \) from equation (2) into this: \[ 2 + 2xy = 1 \Rightarrow 2xy = -1 \Rightarrow xy = -\frac{1}{2}. \]

Step 2: Use the identity for \( x^5 + y^5 \). 
We need to find \( x^5 + y^5 \). Using known algebraic identities, we calculate \( x^5 + y^5 \) based on the values \( x + y = 1 \) and \( xy = -\frac{1}{2} \). After solving, we find that \( N = \frac{19}{2} \).