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anonymous
 5 years ago
x^2 + xy = 1, xy+y^2 = 3, solve for x and y
anonymous
 5 years ago
x^2 + xy = 1, xy+y^2 = 3, solve for x and y

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since you have xy in each equation they must be the same result. Solve the first and second equations for xy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0xy = 1  x^2 and xy = 3  y^2 1  x^2 = 3  y^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Solve for y: subtract 3 from both sides 2  x^2 = y^2 take the opposite of both sides 2 + x^2 = y^2 square root both sides sqrt(x^2 + 2) = y y = sqrt(x^2 + 2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Solve for x: 1  x^2 = 3  y^2 subtract 1 from both sides x^2 = 2  y^2 take the opposite of both sides x^2 = y^2  2 square root both sides x = sqrt(y^2  2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is there another way to solve this?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think the results should be real numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Come on, we can be more elegant than this, guys: Add them both: x^2 + 2xy + y^2 = 4 (x+y)^2 = 4 => x+y = ±2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Case 1: x = 2y gives (2y)y = 3y^2 => y = 3/2, x = 1/2 Case 2 x = (2+y) gives y(2+y) = 3y^2 gives y = 3/2, x = 1/2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's how we do it at Cambridge, featheres!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You're welcome, kiddo.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x=1/2,x=1/2 and y=3/2 and y=3/2 are the answers x(x+y)=1 and y(x+y)=3 implies x/y=1/3 and 3x=y .PLug in this in x^2+xy=1,x^2+3x^2=1,4x^2=1, x^2=1,x=1/2 or 1/2 now we have y3x=3(1/2)=3/2 and 3/2
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