What is the answer to solvefor x,y^2-2xy+1=0 ?

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

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solvefor x,y^2-2xy+1=0

solve for

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

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

Solution steps

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

Move

y^{2}

to the right side

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

Move

1

to the right side

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

Divide both sides by

- 2 y; y \neq = 0

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

lo g_{10} (4) (2 x) = 14

2 + y = \frac{3}{2}

x, (x - 2) 2 = - 14 (y - 2)

a, y = \frac{a}{x ^{2} + b}

4 \frac{1}{5} = 3 p

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