Subtract the two equations:
Substitute in the original equation:
x-4=0; x+4=0; x-8=0; x+8=0
x=4; x=-4; x=8; x=-8
Substitute each in {{y=32/x}}}
;; ; ;; ;
Solutions are
(x,y)=(4,8), (-4,-8), (8,4), (-8,-4)
Edwin
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website! the squarqe of the sum of two numbers is 144 and the sum of their squares is 80. find the number
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(x + y)^2 = 144
x^2 + y^2 = 80
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x + y = 12 ---> y = 12 - x
x^2 + y^2 = 80
x^2 + (12-x)^2 = 80
2x^2 - 24x + 64 = 0
x^2 - 12x + 32 = 0
x = 4, x = 8
y = 8, y = 4
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x + y = -12 ---> y = -12 - x
x^2 + y^2 = 80
x^2 + (-12-x)^2 = 80
2x^2 + 24x + 64 = 0
x^2 + 12x + 32 = 0
x = -4, x = -8
y = -8, y = -4
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One number is + or -4, the other is + or -8,
and the absolute value of the sum is 12.