Solve xy-2xy/x^2+xy*x^2+2xy+y^2/x^2-2xy

$$\frac { x y - 2 x y } { x ^ { 2 } + x y } \cdot \frac { x ^ { 2 } + 2 x y + y ^ { 2 } } { x ^ { 2 } - 2 x y }$$

Evaluate

$$-\frac{y\left(x+y\right)}{x\left(x-2y\right)}$$

Short Solution Steps

Combine xy and -2xy to get -xy.

$$\frac{-xy}{x^{2}+xy}\times \frac{x^{2}+2xy+y^{2}}{x^{2}-2xy}$$

Factor the expressions that are not already factored in \frac{-xy}{x^{2}+xy}.

$$\frac{-xy}{x\left(x+y\right)}\times \frac{x^{2}+2xy+y^{2}}{x^{2}-2xy}$$

Cancel out x in both numerator and denominator.

$$\frac{-y}{x+y}\times \frac{x^{2}+2xy+y^{2}}{x^{2}-2xy}$$

Multiply \frac{-y}{x+y} times \frac{x^{2}+2xy+y^{2}}{x^{2}-2xy} by multiplying numerator times numerator and denominator times denominator.

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

Factor the expressions that are not already factored.

$$\frac{-y\left(x+y\right)^{2}}{x\left(x+y\right)\left(x-2y\right)}$$

Cancel out x+y in both numerator and denominator.

$$\frac{-y\left(x+y\right)}{x\left(x-2y\right)}$$

Expand the expression.

$$\frac{-xy-y^{2}}{x^{2}-2xy}$$

Expand

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

Short Solution Steps

Combine xy and -2xy to get -xy.

$$\frac{-xy}{x^{2}+xy}\times \frac{x^{2}+2xy+y^{2}}{x^{2}-2xy}$$

Factor the expressions that are not already factored in \frac{-xy}{x^{2}+xy}.

$$\frac{-xy}{x\left(x+y\right)}\times \frac{x^{2}+2xy+y^{2}}{x^{2}-2xy}$$

Cancel out x in both numerator and denominator.

$$\frac{-y}{x+y}\times \frac{x^{2}+2xy+y^{2}}{x^{2}-2xy}$$

Multiply \frac{-y}{x+y} times \frac{x^{2}+2xy+y^{2}}{x^{2}-2xy} by multiplying numerator times numerator and denominator times denominator.

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

Factor the expressions that are not already factored.

$$\frac{-y\left(x+y\right)^{2}}{x\left(x+y\right)\left(x-2y\right)}$$

Cancel out x+y in both numerator and denominator.

$$\frac{-y\left(x+y\right)}{x\left(x-2y\right)}$$

Expand the expression.

$$\frac{-xy-y^{2}}{x^{2}-2xy}$$