x^2 + xy = 1, xy+y^2 = 3, solve for x and y

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x^2 + xy = 1, xy+y^2 = 3, solve for x and y

Mathematics
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Since you have xy in each equation they must be the same result. Solve the first and second equations for xy
xy = 1 - x^2 and xy = 3 - y^2 1 - x^2 = 3 - y^2
Solve 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)

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Solve 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)
is there another way to solve this?
i think the results should be real numbers
Come 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
Case 1: x = 2-y gives (2-y)y = 3-y^2 => y = 3/2, x = 1/2 Case 2 x = -(2+y) gives -y(2+y) = 3-y^2 gives y = -3/2, x = -1/2
That's how we do it at Cambridge, featheres!
thanks
You're welcome, kiddo.
x=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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