$$\frac { \frac { 1 } { a + 1 } + \frac { 2 a } { 1 - a } } { \frac { 1 + a } { 1 - a } - \frac { 2 a } { 1 + a } }$$
Evaluate
$\frac{2a^{2}+a+1}{3a^{2}+1}$
Short Solution Steps
To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $a+1$ and $1-a$ is $\left(a+1\right)\left(-a+1\right)$. Multiply $\frac{1}{a+1}$ times $\frac{-a+1}{-a+1}$. Multiply $\frac{2a}{1-a}$ times $\frac{a+1}{a+1}$.
Since $\frac{-a+1}{\left(a+1\right)\left(-a+1\right)}$ and $\frac{2a\left(a+1\right)}{\left(a+1\right)\left(-a+1\right)}$ have the same denominator, add them by adding their numerators.
To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $1-a$ and $1+a$ is $\left(a+1\right)\left(-a+1\right)$. Multiply $\frac{1+a}{1-a}$ times $\frac{a+1}{a+1}$. Multiply $\frac{2a}{1+a}$ times $\frac{-a+1}{-a+1}$.
Since $\frac{\left(1+a\right)\left(a+1\right)}{\left(a+1\right)\left(-a+1\right)}$ and $\frac{2a\left(-a+1\right)}{\left(a+1\right)\left(-a+1\right)}$ have the same denominator, subtract them by subtracting their numerators.
Divide $\frac{a+1+2a^{2}}{\left(a+1\right)\left(-a+1\right)}$ by $\frac{1+3a^{2}}{\left(a+1\right)\left(-a+1\right)}$ by multiplying $\frac{a+1+2a^{2}}{\left(a+1\right)\left(-a+1\right)}$ by the reciprocal of $\frac{1+3a^{2}}{\left(a+1\right)\left(-a+1\right)}$.
To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $a+1$ and $1-a$ is $\left(a+1\right)\left(-a+1\right)$. Multiply $\frac{1}{a+1}$ times $\frac{-a+1}{-a+1}$. Multiply $\frac{2a}{1-a}$ times $\frac{a+1}{a+1}$.
Since $\frac{-a+1}{\left(a+1\right)\left(-a+1\right)}$ and $\frac{2a\left(a+1\right)}{\left(a+1\right)\left(-a+1\right)}$ have the same denominator, add them by adding their numerators.
To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $1-a$ and $1+a$ is $\left(a+1\right)\left(-a+1\right)$. Multiply $\frac{1+a}{1-a}$ times $\frac{a+1}{a+1}$. Multiply $\frac{2a}{1+a}$ times $\frac{-a+1}{-a+1}$.
Since $\frac{\left(1+a\right)\left(a+1\right)}{\left(a+1\right)\left(-a+1\right)}$ and $\frac{2a\left(-a+1\right)}{\left(a+1\right)\left(-a+1\right)}$ have the same denominator, subtract them by subtracting their numerators.
Divide $\frac{a+1+2a^{2}}{\left(a+1\right)\left(-a+1\right)}$ by $\frac{1+3a^{2}}{\left(a+1\right)\left(-a+1\right)}$ by multiplying $\frac{a+1+2a^{2}}{\left(a+1\right)\left(-a+1\right)}$ by the reciprocal of $\frac{1+3a^{2}}{\left(a+1\right)\left(-a+1\right)}$.
