Question

If 4x^2+4y^2-20x-12y+34=0, then find thevalue of xy pls answer fast. with statements

Answer 1

Step-by-step explanation:

4x² + 4y² - 20x -12y +34=0

4x² + 4y² - 20x - 12y = -34

4x² + 4y² - 20x - y = -34/12

4x² + 4y² - x - y = -34/12 ÷ 20

4x² + 4y² - x- y = -34/12 * 1/20

4x² + 4y² - x - y = -17/120

x² + 4y² - x - y = -17/120 ÷ 4

x² + y² - x - y = -17/480 ÷ 4

x² + y² - x - y = 1920

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