Solve p-q/p+q*frac{p^{2}-q^{2}}{2p-q}divp^2-2pq+q^2/4p^2-q^2

$$\frac { p - q } { p + q } \times \frac { p ^ { 2 } - q ^ { 2 } } { 2 p - q } \div \frac { p ^ { 2 } - 2 p q + q ^ { 2 } } { 4 p ^ { 2 } - q ^ { 2 } }$$

Evaluate

$$2p+q$$

Short Solution Steps

Multiply \frac{p-q}{p+q} times \frac{p^{2}-q^{2}}{2p-q} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{\left(p-q\right)\left(p^{2}-q^{2}\right)}{\left(p+q\right)\left(2p-q\right)}}{\frac{p^{2}-2pq+q^{2}}{4p^{2}-q^{2}}}$$

Divide \frac{\left(p-q\right)\left(p^{2}-q^{2}\right)}{\left(p+q\right)\left(2p-q\right)} by \frac{p^{2}-2pq+q^{2}}{4p^{2}-q^{2}} by multiplying \frac{\left(p-q\right)\left(p^{2}-q^{2}\right)}{\left(p+q\right)\left(2p-q\right)} by the reciprocal of \frac{p^{2}-2pq+q^{2}}{4p^{2}-q^{2}}.

$$\frac{\left(p-q\right)\left(p^{2}-q^{2}\right)\left(4p^{2}-q^{2}\right)}{\left(p+q\right)\left(2p-q\right)\left(p^{2}-2pq+q^{2}\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(p+q\right)\left(2p+q\right)\left(2p-q\right)\left(p-q\right)^{2}}{\left(p+q\right)\left(2p-q\right)\left(p-q\right)^{2}}$$

Cancel out \left(p+q\right)\left(2p-q\right)\left(p-q\right)^{2} in both numerator and denominator.

$$2p+q$$

Expand

$$2p+q$$

Short Solution Steps

Multiply \frac{p-q}{p+q} times \frac{p^{2}-q^{2}}{2p-q} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\frac{\left(p-q\right)\left(p^{2}-q^{2}\right)}{\left(p+q\right)\left(2p-q\right)}}{\frac{p^{2}-2pq+q^{2}}{4p^{2}-q^{2}}}$$

Divide \frac{\left(p-q\right)\left(p^{2}-q^{2}\right)}{\left(p+q\right)\left(2p-q\right)} by \frac{p^{2}-2pq+q^{2}}{4p^{2}-q^{2}} by multiplying \frac{\left(p-q\right)\left(p^{2}-q^{2}\right)}{\left(p+q\right)\left(2p-q\right)} by the reciprocal of \frac{p^{2}-2pq+q^{2}}{4p^{2}-q^{2}}.

$$\frac{\left(p-q\right)\left(p^{2}-q^{2}\right)\left(4p^{2}-q^{2}\right)}{\left(p+q\right)\left(2p-q\right)\left(p^{2}-2pq+q^{2}\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(p+q\right)\left(2p+q\right)\left(2p-q\right)\left(p-q\right)^{2}}{\left(p+q\right)\left(2p-q\right)\left(p-q\right)^{2}}$$

Cancel out \left(p+q\right)\left(2p-q\right)\left(p-q\right)^{2} in both numerator and denominator.

$$2p+q$$