#1**+1 **

In order to simplify this expression, we must create a common denominator. Otherwise, subtracting complicated fractions is not feasible. Currently, \(\frac{a^2+1}{a^2-1}-\frac{a-1}{a+1}\) does not have this, so we must create one.

The denominator of the first term, \(a^2-1\) can be factored because it is a difference of two squares. Sometimes, factoring the denominator fully is helpful since it can make identifying the least common multiple all that more painless. In this case, \(a^2-1=(a+1)(a-1)\).

We now have \(\frac{a^2+1}{(a+1)(a-1)}-\frac{a-1}{a+1}\). Well, would you look at that! It is clear here that \((a+1)(a-1)\) is the least common multiple of the denominators since \(a+1\) is a factor of the product of binomials. Let's manipulate \(\frac{a-1}{a+1}\) so that we can create the desired common denominator.

\(\frac{a-1}{a+1}\) | As aforementioned, let's convert this into a fraction with a common denominator. |

\(\frac{a-1}{a+1}*\frac{a-1}{a-1}\) | Of course, the actual value of the fraction remains unchanged since I am really multiplying the fraction by one. Now, let's combine. |

\(\frac{(a-1)(a-1)}{(a+1)(a-1)}\) | |

Ok, we have now changed the problem from \(\frac{a^2+1}{a^2-1}-\frac{a-1}{a+1}\) to \(\frac{a^2+1}{(a+1)(a-1)}-\frac{(a-1)(a-1)}{(a+1)(a-1)}\). This transformation is indeed a positive one because we can now combine the fractions. We have now overcome the first stumbling block.

\(\frac{a^2+1}{(a+1)(a-1)}-\frac{(a-1)(a-1)}{(a+1)(a-1)}\) | Combine the fractions now that the same denominator of both terms exist. |

\(\frac{a^2+1-(a-1)(a-1)}{(a+1)(a-1)}\) | Let's determine the product of the binomials. Since both binomials are identical, we can utilize the rule that \((x-y)^2=x^2-2xy+y^2\) |

\(\frac{a^2+1-(a^2-2a+1)}{(a+1)(a-1)}\) | Distribute the negative sign to all the terms it contains in parentheses. |

\(\frac{a^2+1-a^2+2a-1}{(a+1)(a-1)}\) | Look at that! In the numerator, the a^{2}-term will cancel out and so will the constants! |

\(\frac{2a}{(a+1)(a-1)}\) | We can do the multiplication of the binomials in the denominator again. We already know what the product is. |

\(\frac{2a}{a^2-1}\) | Of course, let's not forget the restrictions for the value of a. |

\(\frac{2a}{a^2-1}, a\neq\pm1\) | If a was equal to the restricted values, a division by zero would occur, which is forbidden. |

TheXSquaredFactor
Jan 12, 2018

#1**+1 **

Best Answer

In order to simplify this expression, we must create a common denominator. Otherwise, subtracting complicated fractions is not feasible. Currently, \(\frac{a^2+1}{a^2-1}-\frac{a-1}{a+1}\) does not have this, so we must create one.

The denominator of the first term, \(a^2-1\) can be factored because it is a difference of two squares. Sometimes, factoring the denominator fully is helpful since it can make identifying the least common multiple all that more painless. In this case, \(a^2-1=(a+1)(a-1)\).

We now have \(\frac{a^2+1}{(a+1)(a-1)}-\frac{a-1}{a+1}\). Well, would you look at that! It is clear here that \((a+1)(a-1)\) is the least common multiple of the denominators since \(a+1\) is a factor of the product of binomials. Let's manipulate \(\frac{a-1}{a+1}\) so that we can create the desired common denominator.

\(\frac{a-1}{a+1}\) | As aforementioned, let's convert this into a fraction with a common denominator. |

\(\frac{a-1}{a+1}*\frac{a-1}{a-1}\) | Of course, the actual value of the fraction remains unchanged since I am really multiplying the fraction by one. Now, let's combine. |

\(\frac{(a-1)(a-1)}{(a+1)(a-1)}\) | |

Ok, we have now changed the problem from \(\frac{a^2+1}{a^2-1}-\frac{a-1}{a+1}\) to \(\frac{a^2+1}{(a+1)(a-1)}-\frac{(a-1)(a-1)}{(a+1)(a-1)}\). This transformation is indeed a positive one because we can now combine the fractions. We have now overcome the first stumbling block.

\(\frac{a^2+1}{(a+1)(a-1)}-\frac{(a-1)(a-1)}{(a+1)(a-1)}\) | Combine the fractions now that the same denominator of both terms exist. |

\(\frac{a^2+1-(a-1)(a-1)}{(a+1)(a-1)}\) | Let's determine the product of the binomials. Since both binomials are identical, we can utilize the rule that \((x-y)^2=x^2-2xy+y^2\) |

\(\frac{a^2+1-(a^2-2a+1)}{(a+1)(a-1)}\) | Distribute the negative sign to all the terms it contains in parentheses. |

\(\frac{a^2+1-a^2+2a-1}{(a+1)(a-1)}\) | Look at that! In the numerator, the a^{2}-term will cancel out and so will the constants! |

\(\frac{2a}{(a+1)(a-1)}\) | We can do the multiplication of the binomials in the denominator again. We already know what the product is. |

\(\frac{2a}{a^2-1}\) | Of course, let's not forget the restrictions for the value of a. |

\(\frac{2a}{a^2-1}, a\neq\pm1\) | If a was equal to the restricted values, a division by zero would occur, which is forbidden. |

TheXSquaredFactor
Jan 12, 2018