Solve (1-b/1+b)-(1-a/1+a)/1+(frac{1-a{1-b})*(1-b/1+b)}

$$\frac { ( \frac { 1 - b } { 1 + b } ) - ( \frac { 1 - a } { 1 + a } ) } { 1 + ( \frac { 1 - a } { 1 - b } ) \cdot ( \frac { 1 - b } { 1 + b } ) }$$

Evaluate

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

Short Solution Steps

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

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

Since \frac{\left(1-b\right)\left(a+1\right)}{\left(a+1\right)\left(b+1\right)} and \frac{\left(1-a\right)\left(b+1\right)}{\left(a+1\right)\left(b+1\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(1-b\right)\left(a+1\right)-\left(1-a\right)\left(b+1\right).

$$\frac{\frac{1+a-ba-b-b-1+ab+a}{\left(a+1\right)\left(b+1\right)}}{1+\frac{1-a}{1-b}\times \frac{1-b}{1+b}}$$

Combine like terms in 1+a-ba-b-b-1+ab+a.

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

Multiply \frac{1-a}{1-b} times \frac{1-b}{1+b} by multiplying numerator times numerator and denominator times denominator.

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

Cancel out -b+1 in both numerator and denominator.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{b+1}{b+1}.

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

Since \frac{b+1}{b+1} and \frac{-a+1}{b+1} have the same denominator, add them by adding their numerators.

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

Combine like terms in b+1-a+1.

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

Divide \frac{2a-2b}{\left(a+1\right)\left(b+1\right)} by \frac{b+2-a}{b+1} by multiplying \frac{2a-2b}{\left(a+1\right)\left(b+1\right)} by the reciprocal of \frac{b+2-a}{b+1}.

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

Cancel out b+1 in both numerator and denominator.

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

Apply the distributive property by multiplying each term of a+1 by each term of -a+b+2.

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

Combine 2a and -a to get a.

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

Expand

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

Short Solution Steps

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

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

Since \frac{\left(1-b\right)\left(a+1\right)}{\left(a+1\right)\left(b+1\right)} and \frac{\left(1-a\right)\left(b+1\right)}{\left(a+1\right)\left(b+1\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(1-b\right)\left(a+1\right)-\left(1-a\right)\left(b+1\right).

$$\frac{\frac{1+a-ba-b-b-1+ab+a}{\left(a+1\right)\left(b+1\right)}}{1+\frac{1-a}{1-b}\times \frac{1-b}{1+b}}$$

Combine like terms in 1+a-ba-b-b-1+ab+a.

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

Multiply \frac{1-a}{1-b} times \frac{1-b}{1+b} by multiplying numerator times numerator and denominator times denominator.

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

Cancel out -b+1 in both numerator and denominator.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{b+1}{b+1}.

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

Since \frac{b+1}{b+1} and \frac{-a+1}{b+1} have the same denominator, add them by adding their numerators.

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

Combine like terms in b+1-a+1.

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

Divide \frac{2a-2b}{\left(a+1\right)\left(b+1\right)} by \frac{b+2-a}{b+1} by multiplying \frac{2a-2b}{\left(a+1\right)\left(b+1\right)} by the reciprocal of \frac{b+2-a}{b+1}.

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

Cancel out b+1 in both numerator and denominator.

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

Apply the distributive property by multiplying each term of a+1 by each term of -a+b+2.

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

Combine 2a and -a to get a.

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