Solve [2a-3b-2b^2/a(a+1)-b(b+1)+b/a-b]

$$[ \frac { 2 a - 3 b - 2 b ^ { 2 } } { a ( a + 1 ) - b ( b + 1 ) } + \frac { b } { a - b } ]$$

Evaluate

$$\frac{b+2}{a+b+1}$$

Short Solution Steps

Factor a\left(a+1\right)-b\left(b+1\right).

$$\frac{2a-3b-2b^{2}}{\left(a-b\right)\left(a+b+1\right)}+\frac{b}{a-b}$$

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

$$\frac{2a-3b-2b^{2}}{\left(a-b\right)\left(a+b+1\right)}+\frac{b\left(a+b+1\right)}{\left(a-b\right)\left(a+b+1\right)}$$

Since \frac{2a-3b-2b^{2}}{\left(a-b\right)\left(a+b+1\right)} and \frac{b\left(a+b+1\right)}{\left(a-b\right)\left(a+b+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{2a-3b-2b^{2}+b\left(a+b+1\right)}{\left(a-b\right)\left(a+b+1\right)}$$

Do the multiplications in 2a-3b-2b^{2}+b\left(a+b+1\right).

$$\frac{2a-3b-2b^{2}+ba+b^{2}+b}{\left(a-b\right)\left(a+b+1\right)}$$

Combine like terms in 2a-3b-2b^{2}+ba+b^{2}+b.

$$\frac{2a-2b+ba-b^{2}}{\left(a-b\right)\left(a+b+1\right)}$$

Factor the expressions that are not already factored in \frac{2a-2b+ba-b^{2}}{\left(a-b\right)\left(a+b+1\right)}.

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

Cancel out a-b in both numerator and denominator.

$$\frac{b+2}{a+b+1}$$

Expand

$$\frac{b+2}{a+b+1}$$

Short Solution Steps

Factor a\left(a+1\right)-b\left(b+1\right).

$$\frac{2a-3b-2b^{2}}{\left(a-b\right)\left(a+b+1\right)}+\frac{b}{a-b}$$

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

$$\frac{2a-3b-2b^{2}}{\left(a-b\right)\left(a+b+1\right)}+\frac{b\left(a+b+1\right)}{\left(a-b\right)\left(a+b+1\right)}$$

Since \frac{2a-3b-2b^{2}}{\left(a-b\right)\left(a+b+1\right)} and \frac{b\left(a+b+1\right)}{\left(a-b\right)\left(a+b+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{2a-3b-2b^{2}+b\left(a+b+1\right)}{\left(a-b\right)\left(a+b+1\right)}$$

Do the multiplications in 2a-3b-2b^{2}+b\left(a+b+1\right).

$$\frac{2a-3b-2b^{2}+ba+b^{2}+b}{\left(a-b\right)\left(a+b+1\right)}$$

Combine like terms in 2a-3b-2b^{2}+ba+b^{2}+b.

$$\frac{2a-2b+ba-b^{2}}{\left(a-b\right)\left(a+b+1\right)}$$

Factor the expressions that are not already factored in \frac{2a-2b+ba-b^{2}}{\left(a-b\right)\left(a+b+1\right)}.

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

Cancel out a-b in both numerator and denominator.

$$\frac{b+2}{a+b+1}$$