Question

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

Answer 1

Since you have xy in each equation they must be the same result. Solve the first and second equations for xy

Answer 2

xy = 1 - x^2 and xy = 3 - y^2

1 - x^2 = 3 - y^2

Answer 3

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)

Answer 4

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)

Answer 5

is there another way to solve this?

Answer 6

i think the results should be real numbers

Answer 7

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

Answer 8

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

Answer 9

That's how we do it at Cambridge, featheres!

Answer 10

thanks

Answer 11

You're welcome, kiddo.

Answer 12

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