What is the answer to solvefor y,x=y^{-2} ?

The answer to solvefor y,x=y^{-2} is y= 1/(sqrt(x)),y=-1/(sqrt(x));\quad x\ne 0

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solvefor y,x=y^{-2}

solve for

y, x = y^{- 2}

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

Solution steps

x = y^{- 2}

Refine

x = \frac{1}{y ^{2}}

Multiply both sides by

y^{2}

x y^{2} = 1

Solve

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

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

x, \frac{2}{x} - y = 3

\frac{6 x}{( x - 1 ) ^{4}} = 0

c, \frac{b}{c} = a

3 x + 5 = \frac{2}{x}

3 x - \frac{1}{2} - \frac{1}{x} = 0

What is the answer to solvefor y,x=y^{-2} ?
The answer to solvefor y,x=y^{-2} is y= 1/(sqrt(x)),y=-1/(sqrt(x));\quad x\ne 0