What is the answer to solvefor x,x+2xy-y^2=2 ?

The answer to solvefor x,x+2xy-y^2=2 is x=(2+y^2)/(1+2y);\quad y\ne-1/2

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solvefor x,x+2xy-y^2=2

solve for

x, x + 2 x y - y^{2} = 2

x = \frac{2 + y ^{2}}{1 + 2 y}; y \neq = - \frac{1}{2}

Solution steps

x + 2 x y - y^{2} = 2

Move

y^{2}

to the right side

x + 2 x y = 2 + y^{2}

Factor

x + 2 x y : x (1 + 2 y)

x (1 + 2 y) = 2 + y^{2}

Divide both sides by

1 + 2 y; y \neq = - \frac{1}{2}

x = \frac{2 + y ^{2}}{1 + 2 y}; y \neq = - \frac{1}{2}

6 x - 4 = 2

x, - 3 x + y = 0

y, 4 x - 3 y = 15

f = 2 (- 2)^{3} + 6

6 x + 12 = 24

What is the answer to solvefor x,x+2xy-y^2=2 ?
The answer to solvefor x,x+2xy-y^2=2 is x=(2+y^2)/(1+2y);\quad y\ne-1/2