Solve 1/m-n-1/m+n:2/3m-3n

$$\frac { 1 } { m - n } - \frac { 1 } { m + n } : \frac { 2 } { 3 m - 3 n }$$

Evaluate

$$\frac{2n-3n^{2}+2m+6mn-3m^{2}}{2\left(m^{2}-n^{2}\right)}$$

Short Solution Steps

Divide \frac{1}{m+n} by \frac{2}{3m-3n} by multiplying \frac{1}{m+n} by the reciprocal of \frac{2}{3m-3n}.

$$\frac{1}{m-n}-\frac{3m-3n}{\left(m+n\right)\times 2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of m-n and \left(m+n\right)\times 2 is 2\left(m+n\right)\left(m-n\right). Multiply \frac{1}{m-n} times \frac{2\left(m+n\right)}{2\left(m+n\right)}. Multiply \frac{3m-3n}{\left(m+n\right)\times 2} times \frac{m-n}{m-n}.

$$\frac{2\left(m+n\right)}{2\left(m+n\right)\left(m-n\right)}-\frac{\left(3m-3n\right)\left(m-n\right)}{2\left(m+n\right)\left(m-n\right)}$$

Since \frac{2\left(m+n\right)}{2\left(m+n\right)\left(m-n\right)} and \frac{\left(3m-3n\right)\left(m-n\right)}{2\left(m+n\right)\left(m-n\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(m+n\right)-\left(3m-3n\right)\left(m-n\right)}{2\left(m+n\right)\left(m-n\right)}$$

Do the multiplications in 2\left(m+n\right)-\left(3m-3n\right)\left(m-n\right).

$$\frac{2m+2n-3m^{2}+3mn+3nm-3n^{2}}{2\left(m+n\right)\left(m-n\right)}$$

Combine like terms in 2m+2n-3m^{2}+3mn+3nm-3n^{2}.

$$\frac{2m+2n-3m^{2}-3n^{2}+6mn}{2\left(m+n\right)\left(m-n\right)}$$

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

$$\frac{2m+2n-3m^{2}-3n^{2}+6mn}{2m^{2}-2n^{2}}$$

Factor

$$\frac{2n-3n^{2}+2m+6mn-3m^{2}}{2\left(m-n\right)\left(m+n\right)}$$