Best Answer

If:

xy = x2 + y2 + 2xy

Then:

x2 + xy + y2 = 0

Do you want to solve it for x?

x2 + xy + (y/2)2 = (y/2)2 - y2

(x + y/2)2 = y2/4 - y2

x + y/2 = ± √(-3y2/4)

x = -y/2 ± y√(-3) / 2

x = (-y ± yi√3) / 2

Q: If xy equals x2 plus y2 plus 2xy?

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x2 + y2 = x2 - 2xy + y2 + 2xy = (x - y)2 + 2xy = 72 + 2*8 = 49 + 16 = 65 You could, instead, solve the two equations for x and y and substitute, but the above method is simpler.

(x + y)2 = x2 + 2xy + y2 So x2 + y2 = (x + y)2 - 2xy = a2 - 2b Then (x2 + y2)2 = x4 + 2x2y2 + y4 So x4 + y4 = (x2 + y2)2 - 2x2y2 = (a2 - 2b)2 - 2b2 = a4 - 4a2b + 4b2 - 2b2 = a4 - 4a2b + 2b2

2xy + 6 = 4xyz Divide through by 2: xy + 3 = 2xyz 3 = 2xyz - xy = y*(2xz - x) Therefore, y = 3/(2xz - x) provided that 2xz - x is not zero.

Area of a square = side length squared (x+y)2 (x+y)(x+y) x2+xy+yx+y2 Area = x2+2xy+y2

x3+xy-x2y2=x(x2+y-xy2)

Related questions

xy + xy = 2xy

x2 + y2 = x2 - 2xy + y2 + 2xy = (x - y)2 + 2xy = 72 + 2*8 = 49 + 16 = 65 You could, instead, solve the two equations for x and y and substitute, but the above method is simpler.

(x + y)2 = x2 + 2xy + y2 So x2 + y2 = (x + y)2 - 2xy = a2 - 2b Then (x2 + y2)2 = x4 + 2x2y2 + y4 So x4 + y4 = (x2 + y2)2 - 2x2y2 = (a2 - 2b)2 - 2b2 = a4 - 4a2b + 4b2 - 2b2 = a4 - 4a2b + 2b2

That equals 2xy + 3y which factors to y(2x + 3)

-1

xy and yx are identical so you have 2xy = 545 ie xy = 272.5. Possible answers 5 and 54.5, 10.9 and 25 etc

2xy + 6 = 4xyz Divide through by 2: xy + 3 = 2xyz 3 = 2xyz - xy = y*(2xz - x) Therefore, y = 3/(2xz - x) provided that 2xz - x is not zero.

x2 + y2 = (x + y)2 => x2 + y2 = x2 + 2xy + y2 => 2xy = 0 => xy = 0 So, one of x and y must be 0.

( 2xy - 4x ) + ( 8y - 16)2 ( xy - 2x ) + 2 ( 4y - 8 ) Answer

Area of a square = side length squared (x+y)2 (x+y)(x+y) x2+xy+yx+y2 Area = x2+2xy+y2

4x-y2=2xy 2x ? y5 if its plus its 7 xy

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