SOLUTION: Find the real Solutions, if any, of each system. {{{xy = 2}}} {{{x^2 + y^2 = 4}}} The answers for this are (Square Root of 2, Square root of 2) (-Square Root of 2, -Square

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the real Solutions, if any, of each system. {{{xy = 2}}} {{{x^2 + y^2 = 4}}} The answers for this are (Square Root of 2, Square root of 2) (-Square Root of 2, -Square      Log On


   

Question 30022: Find the real Solutions, if any, of each system.
xy+=+2
x%5E2+%2B+y%5E2+=+4
The answers for this are
(Square Root of 2, Square root of 2)
(-Square Root of 2, -Square root of 2)
Thankyou,I have the answers, I just need to know how to go about doing these types of problems.
Answer by sdmmadam@yahoo.com(530) About Me  (Show Source):
You can put this solution on YOUR website!
xy = 2 ----(1)
x^2 + y^2 = 4 ----(2)
Use the standard formula
(x-y)^2 = x^2+y^2 - 2xy
=4 -2X(2) using (2) and (1)
=4-4 = 0
(x-y) ^2 = 0
Therefore taking sqrt
(x-y) = 0 ----(3)
Which means x = y ----(3)
Putting x=y in (1)
xy = 2
yXy = 2
y^2 = 2
y = +or - root(2)
y = +[root(2)] implies from (3) x also = +[root(2)]
y = -[root(2)] implies from (3) x also = -[root(2)]
Answer: x = sqrt2, y = sqrt2 (one set of answers)
x = -sqrt2, y = -sqrt2 (another set of answers)
And both sets hold (on oral verification itself)