Question 403156
x2 + y2 = 8
x - y = 0

find the solutions to the system

x^2+y^2=8

x^2-2xy+y^2=8-2xy

(x-y)^2=8-2xy

substitute (x-y) =0

(0)^2=8-2xy

2xy=8
xy=4
x=4/y
...
x-y =0
4/y - y =0
(4-y^2)/y=0
4-y^2=0
y^2=4
take the square root
y= +/- 2

(2,-2) & (-2,2)

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