What are the solutions to the equation solvefor y,x^2y^2+5xy=14 ?

The solutions to the equation solvefor y,x^2y^2+5xy=14 are y= 2/x ,y=-7/x ;\quad x\ne 0

Find the zeros of solvefor y,x^2y^2+5xy=14

The zeros of solvefor y,x^2y^2+5xy=14 are y= 2/x ,y=-7/x ;\quad x\ne 0

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solve for

y, x^{2} y^{2} + 5 x y = 14

y = \frac{2}{x}, y = - \frac{7}{x}; x \neq = 0

Solution steps

x^{2} y^{2} + 5 x y = 14

Move

14

to the left side

x^{2} y^{2} + 5 x y - 14 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 5 x \pm ( 5 x ) ^{2} - 4 x ^{2} ( - 14 )}{2 x ^{2}}

Simplify

(5 x)^{2} - 4 x^{2} (- 14) : 9 x

y_{1, 2} = \frac{- 5 x \pm 9 x}{2 x ^{2}}; x \neq = 0

Separate the solutions

y_{1} = \frac{- 5 x + 9 x}{2 x ^{2}}, y_{2} = \frac{- 5 x - 9 x}{2 x ^{2}}

y = \frac{- 5 x + 9 x}{2 x ^{2}} : \frac{2}{x}

y = \frac{- 5 x - 9 x}{2 x ^{2}} : - \frac{7}{x}

The solutions to the quadratic equation are:

y = \frac{2}{x}, y = - \frac{7}{x}; x \neq = 0

6 x^{2} + 24 x - 126 = 0

4 = 3 x^{2} - 14 x + 4

- 4 a^{2} - 37 = - 137

t^{2} + t = - 7

x^{2} - 4 x + 3 = x - 1

What are the solutions to the equation solvefor y,x^2y^2+5xy=14 ?
The solutions to the equation solvefor y,x^2y^2+5xy=14 are y= 2/x ,y=-7/x ;\quad x\ne 0
Find the zeros of solvefor y,x^2y^2+5xy=14
The zeros of solvefor y,x^2y^2+5xy=14 are y= 2/x ,y=-7/x ;\quad x\ne 0