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Subtracting Rational Expressions

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(a2 + 1)/(a2-1) - (a-1)/(a+1)

Guest Jan 11, 2018

#1
+1807
+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 a2-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
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#1
+1807
+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 a2-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
#2
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Thank you very much!!

Guest Jan 12, 2018

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