What are the solutions to the equation solvefor x,2x^2y-2xy=0 ?

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

Find the zeros of solvefor x,2x^2y-2xy=0

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

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solve for

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

x = 1, x = 0; y \neq = 0

Solution steps

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

Write in the standard form

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

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 y ) \pm ( - 2 y ) ^{2} - 4 \cdot 2 y \cdot 0}{2 \cdot 2 y}

Simplify

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

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

Separate the solutions

x_{1} = \frac{- ( - 2 y ) + 2 y}{2 \cdot 2 y}, x_{2} = \frac{- ( - 2 y ) - 2 y}{2 \cdot 2 y}

x = \frac{- ( - 2 y ) + 2 y}{2 \cdot 2 y} : 1

x = \frac{- ( - 2 y ) - 2 y}{2 \cdot 2 y} : 0

The solutions to the quadratic equation are:

x = 1, x = 0; y \neq = 0

y, x^{3} + x y^{2} - 2 y^{2} = 0

A x^{2} + 2 x + 3 = x^{2} - B x + C

x^{2} + 10 x + 5 = 30

y, x^{3} + y^{2} = 2 x y^{2}

x (x - 9) = 2 + (2 x + 1) (x - 4)

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