Case 1: x+y ≠ 0
Multiply eqn 1 by 2, set equal to eqn 2.
2x^2 + 6y = 2y^2 - 6x
2(x^2 - y^2) = -6(x+y) (divide by x+y)
x-y = -3
y = x+3 plug this into
2y^2 - 6x = 18
2(x+3)^2 - 6x = 18
2x^2 + 6x = 0
x = {0, -3}
(x,y) = {(0,3),(-3,0)}
But these are only 2 solutions provided x+y cannot be zero
Case 2: x+y = 0
2y^2 - 6x = 18
2(-x)^2 - 6x = 18
2x^2 - 6x - 18 = 0
x^2 - 3x - 9 = 0
x = (3 ± √(45))/2
x = (3/2)(1 ± √5)
(x,y) = {((3/2)(1 + √5),-(3/2)(1 + √5)),((3/2)(1 - √5),-(3/2)(1 - √5))}
sum(y values) = 3 - (3/2)(1 + √5) - (3/2)(1 - √5) = 0
sum(y values) = -3 + ((3/2)(1 + √5) + (3/2)(1 - √5)) = 0
The requested formula calculates to 0.
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Posted by Larry
on 2024-11-18 09:13:17 |
Solution
