How do you express $(\frac{y^{2} - x y}{x^{2} + x y} - x y + y^{2}) (\frac{x}{x - y}) + (\frac{y}{x + y})$ $((y^2-xy)/(x^2+xy)-xy+y^2)(x/(x-y))+(y/(x+y))$ as a single fraction?

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How do you express $(\frac{y^{2} - x y}{x^{2} + x y} - x y + y^{2}) (\frac{x}{x - y}) + (\frac{y}{x + y})$ $((y^2-xy)/(x^2+xy)-xy+y^2)(x/(x-y))+(y/(x+y))$ as a single fraction?

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Answer 1 · 2017-05-26T05:53:52

$(\frac{y^{2} - x y}{x^{2} + x y} - x y + y^{2}) (\frac{x}{x - y}) + (\frac{y}{x + y}) = - x y$ $((y^2-xy)/(x^2+xy)-xy+y^2)(x/(x-y))+(y/(x+y))=-xy$

Explanation:

$(\frac{y^{2} - x y}{x^{2} + x y} - x y + y^{2}) (\frac{x}{x - y}) + (\frac{y}{x + y})$ $((y^2-xy)/(x^2+xy)-xy+y^2)(x/(x-y))+(y/(x+y))$

$= (\frac{- y (x - y)}{x (x + y)} - y (x - y)) (\frac{x}{x - y}) + (\frac{y}{x + y})$ $=((-y(x-y))/(x(x+y))-y(x-y))(x/(x-y))+(y/(x+y))$

$= (\frac{- y (x - y)}{x (x + y)} + \frac{- y (x - y) x (x + y)}{x (x + y)}) (\frac{x}{x - y}) + (\frac{y}{x + y})$ $=((-y(x-y))/(x(x+y))+(-y(x-y)x(x+y))/(x(x+y)))(x/(x-y))+(y/(x+y))$

$= (\frac{- y (x - y) (1 + x (x + y))}{x (x + y)}) (\frac{x}{x - y}) + (\frac{y}{x + y})$ $=((-y(x-y)(1+x(x+y)))/(x(x+y)))(x/(x-y))+(y/(x+y))$

$=((-ycancel((x-y))(1+x(x+y)))/(cancelx(x+y)))(cancelx/cancel((x-y)))+(y/(x+y))$

$=(-y(1+x(x+y)))/((x+y))+(y/(x+y))$

$=-y/((x+y))-(yx(x+y))/((x+y))+(y/(x+y))$

$=-xy$

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How do you express $(\frac{y^{2} - x y}{x^{2} + x y} - x y + y^{2}) (\frac{x}{x - y}) + (\frac{y}{x + y})$ $((y^2-xy)/(x^2+xy)-xy+y^2)(x/(x-y))+(y/(x+y))$ as a single fraction?

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Explanation:

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How do you express ((y^2-xy)/(x^2+xy)-xy+y^2)(x/(x-y))+(y/(x+y))(y2−xyx2+xy−xy+y2)(xx−y)+(yx+y)((y^2-xy)/(x^2+xy)-xy+y^2)(x/(x-y))+(y/(x+y)) as a single fraction?

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Explanation:

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How do you express $(\frac{y^{2} - x y}{x^{2} + x y} - x y + y^{2}) (\frac{x}{x - y}) + (\frac{y}{x + y})$ $((y^2-xy)/(x^2+xy)-xy+y^2)(x/(x-y))+(y/(x+y))$ as a single fraction?