Solve 1/2x-1/x+yleft(x+y/2x-x-yright)

$$\frac{ 1 }{ 2x } - \frac{ 1 }{ x+y } \left( \frac{ x+y }{ 2x } -x-y \right)$$

Evaluate

$$1$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply -x-y times \frac{2x}{2x}.

$$\frac{1}{2x}-\frac{1}{x+y}\left(\frac{x+y}{2x}+\frac{\left(-x-y\right)\times 2x}{2x}\right)$$

Since \frac{x+y}{2x} and \frac{\left(-x-y\right)\times 2x}{2x} have the same denominator, add them by adding their numerators.

$$\frac{1}{2x}-\frac{1}{x+y}\times \frac{x+y+\left(-x-y\right)\times 2x}{2x}$$

Do the multiplications in x+y+\left(-x-y\right)\times 2x.

$$\frac{1}{2x}-\frac{1}{x+y}\times \frac{x+y-2x^{2}-2yx}{2x}$$

Multiply \frac{1}{x+y} times \frac{x+y-2x^{2}-2yx}{2x} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{2x}-\frac{x+y-2x^{2}-2yx}{\left(x+y\right)\times 2x}$$

Factor the expressions that are not already factored in \frac{x+y-2x^{2}-2yx}{\left(x+y\right)\times 2x}.

$$\frac{1}{2x}-\frac{\left(-2x+1\right)\left(x+y\right)}{2x\left(x+y\right)}$$

Cancel out x+y in both numerator and denominator.

$$\frac{1}{2x}-\frac{-2x+1}{2x}$$

Since \frac{1}{2x} and \frac{-2x+1}{2x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1-\left(-2x+1\right)}{2x}$$

Do the multiplications in 1-\left(-2x+1\right).

$$\frac{1+2x-1}{2x}$$

Combine like terms in 1+2x-1.

$$\frac{2x}{2x}$$

Cancel out 2x in both numerator and denominator.

$$1$$

Factor

$$1$$