SOLUTION: Solve y=1/x and x^2+y^2=2 simultaneously. Begin by subtracting the equation of the hyperbola from the equation of the circle.

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Question 1178359: Solve y=1/x and x^2+y^2=2 simultaneously.
Begin by subtracting the equation of the hyperbola from the equation of the circle.

Found 2 solutions by math_helper, ikleyn:
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!

I can solve it this way (substitution):
y=1%2Fx
x%5E2%2By%5E2+=+2+
Substitute 1/x for y in the 2nd eqn:
+x%5E2+%2B+%281%2Fx%29%5E2+=+2+
Simplifies to:
+x%5E4+-+2x%5E2+%2B+1+=+0+
Factors to:
+%28x%5E2+-+1%29+=+0+
Solutions:
+x%5E2+=+1+ --> x = -1 and/or x = 1
For x=-1 we get y=-1 and x%5E2%2By%5E2+=+%28-1%29%5E2%2B%28-1%29%5E2+=+1%2B1+=+2+
For x=1 we get y=1 and x%5E2%2By%5E2+=+%281%29%5E2+%2B+%281%29%5E2+=+1%2B1+=+2+
Both solutions (1,1) and (-1,-1) satisify the equations.

Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.

Our starting equations are


    x^2 + y^2 = 2      (1)

    y = 1%2Fx             (2)


From equation (2),  

    xy = 1             (3)



So, I will multiply equation (3) by 2 (both sides) and then subtract it from equation (1).  I will get then


    x^2 - 2xy + y^2 = 0,

or

    (x-y)^2 = 0.


It means  x = y,  and then from equation (1) I have


    2x^2 = 2,  x^2 = 1, x = +/- 1.


Thus the two solutions are  

    x = y = 1   and/or   x = y = -1.

Solved.