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

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

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solvefor x,p^2+p(2x-y)-2xy=0

solve for

x, p^{2} + p (2 x - y) - 2 x y = 0

x = - \frac{p}{2}; p \neq = y

Solution steps

p^{2} + p (2 x - y) - 2 x y = 0

Move

p^{2}

to the right side

p (2 x - y) - 2 x y = - p^{2}

Expand

p (2 x - y) : 2 p x - p y

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

Move

p y

to the right side

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

Factor

2 p x - 2 x y : 2 x (p - y)

2 x (p - y) = - p^{2} + p y

Divide both sides by

2 (p - y); p \neq = y

x = - \frac{p}{2}; p \neq = y

x, x - 7 = 7 y

\frac{4 u - 7}{- 7} = \frac{u - 4}{4}

x, y^{0.12} = 6 x + 3 y

x, 5 x + 3 y = 95

\frac{x}{2 + 1} = x

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