Solve m/m+n-m/2m-n

$$\frac { m } { m + n } - \frac { m } { 2 m - n }$$

Evaluate

$$\frac{m\left(m-2n\right)}{\left(2m-n\right)\left(m+n\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of m+n and 2m-n is \left(m+n\right)\left(2m-n\right). Multiply \frac{m}{m+n} times \frac{2m-n}{2m-n}. Multiply \frac{m}{2m-n} times \frac{m+n}{m+n}.

$$\frac{m\left(2m-n\right)}{\left(m+n\right)\left(2m-n\right)}-\frac{m\left(m+n\right)}{\left(m+n\right)\left(2m-n\right)}$$

Since \frac{m\left(2m-n\right)}{\left(m+n\right)\left(2m-n\right)} and \frac{m\left(m+n\right)}{\left(m+n\right)\left(2m-n\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{m\left(2m-n\right)-m\left(m+n\right)}{\left(m+n\right)\left(2m-n\right)}$$

Do the multiplications in m\left(2m-n\right)-m\left(m+n\right).

$$\frac{2m^{2}-mn-m^{2}-mn}{\left(m+n\right)\left(2m-n\right)}$$

Combine like terms in 2m^{2}-mn-m^{2}-mn.

$$\frac{m^{2}-2mn}{\left(m+n\right)\left(2m-n\right)}$$

Expand \left(m+n\right)\left(2m-n\right).

$$\frac{m^{2}-2mn}{2m^{2}+mn-n^{2}}$$

Factor

$$\frac{m\left(m-2n\right)}{\left(2m-n\right)\left(m+n\right)}$$