What are the solutions to the equation solvefor y,x^2y^2+xy=2 ?

The solutions to the equation solvefor y,x^2y^2+xy=2 are y= 1/x ,y=-2/x ;\quad x\ne 0

Find the zeros of solvefor y,x^2y^2+xy=2

The zeros of solvefor y,x^2y^2+xy=2 are y= 1/x ,y=-2/x ;\quad x\ne 0

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solve for

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

y = \frac{1}{x}, y = - \frac{2}{x}; x \neq = 0

Solution steps

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

Move

2

to the left side

x^{2} y^{2} + x y - 2 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- x \pm x ^{2} - 4 x ^{2} ( - 2 )}{2 x ^{2}}

Simplify

x^{2} - 4 x^{2} (- 2) : 3 x

y_{1, 2} = \frac{- x \pm 3 x}{2 x ^{2}}; x \neq = 0

Separate the solutions

y_{1} = \frac{- x + 3 x}{2 x ^{2}}, y_{2} = \frac{- x - 3 x}{2 x ^{2}}

y = \frac{- x + 3 x}{2 x ^{2}} : \frac{1}{x}

y = \frac{- x - 3 x}{2 x ^{2}} : - \frac{2}{x}

The solutions to the quadratic equation are:

y = \frac{1}{x}, y = - \frac{2}{x}; x \neq = 0

(x + 2)^{2} + 25 = 50

x, x^{2} = 4 p (x) y

y, x^{2 x + 3} = y^{2} + 1

0 = 2 x^{2} + 6 x - 14

4 (x + 2)^{2} + 4 = 40

What are the solutions to the equation solvefor y,x^2y^2+xy=2 ?
The solutions to the equation solvefor y,x^2y^2+xy=2 are y= 1/x ,y=-2/x ;\quad x\ne 0
Find the zeros of solvefor y,x^2y^2+xy=2
The zeros of solvefor y,x^2y^2+xy=2 are y= 1/x ,y=-2/x ;\quad x\ne 0