What are the solutions to the equation solvefor y,x^2y^2+xy=12 ?

The solutions to the equation solvefor y,x^2y^2+xy=12 are y= 3/x ,y=-4/x ;\quad x\ne 0

Find the zeros of solvefor y,x^2y^2+xy=12

The zeros of solvefor y,x^2y^2+xy=12 are y= 3/x ,y=-4/x ;\quad x\ne 0

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solve for

y, x^{2} y^{2} + x y = 12

y = \frac{3}{x}, y = - \frac{4}{x}; x \neq = 0

Solution steps

x^{2} y^{2} + x y = 12

Move

12

to the left side

x^{2} y^{2} + x y - 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- x \pm x ^{2} - 4 x ^{2} ( - 12 )}{2 x ^{2}}

Simplify

x^{2} - 4 x^{2} (- 12) : 7 x

y_{1, 2} = \frac{- x \pm 7 x}{2 x ^{2}}; x \neq = 0

Separate the solutions

y_{1} = \frac{- x + 7 x}{2 x ^{2}}, y_{2} = \frac{- x - 7 x}{2 x ^{2}}

y = \frac{- x + 7 x}{2 x ^{2}} : \frac{3}{x}

y = \frac{- x - 7 x}{2 x ^{2}} : - \frac{4}{x}

The solutions to the quadratic equation are:

y = \frac{3}{x}, y = - \frac{4}{x}; x \neq = 0

3 y^{2} + 7 y - 10 = 0

30 + 13 x - 3 x^{2} = 0

4 (45 + x^{2}) = 196

x (x + 4) 2

2 x^{2} - 3 x = - x^{2} + 3 x

What are the solutions to the equation solvefor y,x^2y^2+xy=12 ?
The solutions to the equation solvefor y,x^2y^2+xy=12 are y= 3/x ,y=-4/x ;\quad x\ne 0
Find the zeros of solvefor y,x^2y^2+xy=12
The zeros of solvefor y,x^2y^2+xy=12 are y= 3/x ,y=-4/x ;\quad x\ne 0