What are the solutions to the equation solvefor x,x^2y^2+1=2xy ?

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

Find the zeros of solvefor x,x^2y^2+1=2xy

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

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solve for

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

x = \frac{1}{y}; y \neq = 0

Solution steps

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

Move

2 x y

to the left side

x^{2} y^{2} + 1 - 2 x y = 0

Write in the standard form

a x^{2} + b x + c = 0

y^{2} x^{2} - 2 y x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 y ) \pm ( - 2 y ) ^{2} - 4 y ^{2} \cdot 1}{2 y ^{2}}

(- 2 y)^{2} - 4 y^{2} \cdot 1 = 0

x_{1, 2} = \frac{- ( - 2 y ) \pm 0}{2 y ^{2}}; y \neq = 0

x = \frac{- ( - 2 y )}{2 y ^{2}}; y \neq = 0

Simplify

\frac{- ( - 2 y )}{2 y ^{2}} : \frac{1}{y}

x = \frac{1}{y}; y \neq = 0

The solution to the quadratic equation is:

x = \frac{1}{y}; y \neq = 0

29 (x + 1) + 3 = - x^{2} + 27 x + 6

\frac{350 x ^{2}}{2} = 6.64 \cdot 1257 \cdot (540 - x)

2 (3 h - 1) = 4 - 2 h (h + 5)

(2 x - 3)^{2} = (x + 1) (2 x - 3)

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

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