Solve (2a-b/a+b+b/a-b)diva-2b/a-b

$$( \frac { 2 a - b } { a + b } + \frac { b } { a - b } ) \div \frac { a - 2 b } { a - b }$$

Evaluate

$$\frac{2\left(a^{2}-ab+b^{2}\right)}{\left(a-2b\right)\left(a+b\right)}$$

Short Solution Steps

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

$$\frac{\frac{\left(2a-b\right)\left(a-b\right)}{\left(a+b\right)\left(a-b\right)}+\frac{b\left(a+b\right)}{\left(a+b\right)\left(a-b\right)}}{\frac{a-2b}{a-b}}$$

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

$$\frac{\frac{\left(2a-b\right)\left(a-b\right)+b\left(a+b\right)}{\left(a+b\right)\left(a-b\right)}}{\frac{a-2b}{a-b}}$$

Do the multiplications in \left(2a-b\right)\left(a-b\right)+b\left(a+b\right).

$$\frac{\frac{2a^{2}-2ab-ba+b^{2}+ba+b^{2}}{\left(a+b\right)\left(a-b\right)}}{\frac{a-2b}{a-b}}$$

Combine like terms in 2a^{2}-2ab-ba+b^{2}+ba+b^{2}.

$$\frac{\frac{2a^{2}+2b^{2}-2ab}{\left(a+b\right)\left(a-b\right)}}{\frac{a-2b}{a-b}}$$

Divide \frac{2a^{2}+2b^{2}-2ab}{\left(a+b\right)\left(a-b\right)} by \frac{a-2b}{a-b} by multiplying \frac{2a^{2}+2b^{2}-2ab}{\left(a+b\right)\left(a-b\right)} by the reciprocal of \frac{a-2b}{a-b}.

$$\frac{\left(2a^{2}+2b^{2}-2ab\right)\left(a-b\right)}{\left(a+b\right)\left(a-b\right)\left(a-2b\right)}$$

Cancel out a-b in both numerator and denominator.

$$\frac{2a^{2}-2ab+2b^{2}}{\left(a+b\right)\left(a-2b\right)}$$

Apply the distributive property by multiplying each term of a+b by each term of a-2b.

$$\frac{2a^{2}-2ab+2b^{2}}{a^{2}-2ab+ba-2b^{2}}$$

Combine -2ab and ba to get -ab.

$$\frac{2a^{2}-2ab+2b^{2}}{a^{2}-ab-2b^{2}}$$

Expand

$$\frac{2\left(a^{2}-ab+b^{2}\right)}{\left(a-2b\right)\left(a+b\right)}$$

Short Solution Steps

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

$$\frac{\frac{\left(2a-b\right)\left(a-b\right)}{\left(a+b\right)\left(a-b\right)}+\frac{b\left(a+b\right)}{\left(a+b\right)\left(a-b\right)}}{\frac{a-2b}{a-b}}$$

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

$$\frac{\frac{\left(2a-b\right)\left(a-b\right)+b\left(a+b\right)}{\left(a+b\right)\left(a-b\right)}}{\frac{a-2b}{a-b}}$$

Do the multiplications in \left(2a-b\right)\left(a-b\right)+b\left(a+b\right).

$$\frac{\frac{2a^{2}-2ab-ba+b^{2}+ba+b^{2}}{\left(a+b\right)\left(a-b\right)}}{\frac{a-2b}{a-b}}$$

Combine like terms in 2a^{2}-2ab-ba+b^{2}+ba+b^{2}.

$$\frac{\frac{2a^{2}+2b^{2}-2ab}{\left(a+b\right)\left(a-b\right)}}{\frac{a-2b}{a-b}}$$

Divide \frac{2a^{2}+2b^{2}-2ab}{\left(a+b\right)\left(a-b\right)} by \frac{a-2b}{a-b} by multiplying \frac{2a^{2}+2b^{2}-2ab}{\left(a+b\right)\left(a-b\right)} by the reciprocal of \frac{a-2b}{a-b}.

$$\frac{\left(2a^{2}+2b^{2}-2ab\right)\left(a-b\right)}{\left(a+b\right)\left(a-b\right)\left(a-2b\right)}$$

Cancel out a-b in both numerator and denominator.

$$\frac{2a^{2}-2ab+2b^{2}}{\left(a+b\right)\left(a-2b\right)}$$

Apply the distributive property by multiplying each term of a+b by each term of a-2b.

$$\frac{2a^{2}-2ab+2b^{2}}{a^{2}-2ab+ba-2b^{2}}$$

Combine -2ab and ba to get -ab.

$$\frac{2a^{2}-2ab+2b^{2}}{a^{2}-ab-2b^{2}}$$