Rational Expression - Stumpeded :/

Nov 2009
21
0
California, United States
Hello, everybody. Currently, I'm learning how to simplify rational expressions. I have the gist of it all, but there's one particular problem that's stumped me. Here it is:

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

The answer which I've gotten is the following:

\(\displaystyle
\frac {(a^2+1)^2}{(-1)(a-1)(a+1)}
\)

However, I checked the answer on Mathway.com and it says it's the following:

\(\displaystyle
\frac {-a^3+a^2+1}{-a+1}
\)

I'm not sure why I'm not getting the answer correct, as I feel I'm doing all the right steps, but surely someone here would know much better than I would. :] I appreciate any help very much.

Colton
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
\(\displaystyle \frac {\frac {a}{1 - a} + \frac {1 + a}{a}}{ \frac {1 - a}{a} + \frac{a}{1 + a}}\)

\(\displaystyle = \frac{\frac{a^2}{a(1 - a)} + \frac{(1 - a)(1 + a)}{a(1 - a)}}{\frac{(1 - a)(1 + a)}{a(1 + a)} + \frac{a^2}{a(1 + a)}}\)

\(\displaystyle = \frac{\frac{a^2 + (1 - a)(1 + a)}{a(1 - a)}}{\frac{(1 - a)(1 + a) + a^2}{a(1 + a)}}\)

\(\displaystyle = \frac{\frac{a^2 + 1 - a^2}{a(1 - a)}}{\frac{1 - a^2 + a^2}{a(1 + a)}}\)

\(\displaystyle = \frac{\frac{1}{a(1 - a)}}{\frac{1}{a(1 + a)}}\)

\(\displaystyle = \frac{a(1 + a)}{a(1 - a)}\)

\(\displaystyle = \frac{1 + a}{1 - a}\).


I do not get either of the answers you supplied...
 
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Nov 2009
21
0
California, United States
Prove It,

Thank you very much for your post and your help. I realized just by looking at your step-by-step process that I screwed up in the very beginning of the equation. Instead of finding a least common denominator between the fractions on the numerator of the overall expression, I used the second fraction's numerator and multiplied it by the fraction on the left... just a silly goof. I appreciate your help. :]

Colton
 
Nov 2009
21
0
California, United States
May I please ask for the step-by-step of one more? O:] Thank you very much, I would greatly appreciate it... I apologize... I'm at the end of the chapter almost and they're getting progressively more difficult for me...

\(\displaystyle

\frac {\frac{1}{a^2} + \frac{2}{ab} + \frac{1}{b^2}}{\frac{1}{a^2} - \frac{1}{b^2}}

\)

Once again, sorry for posting more than one.... this one's got me just as stumped as the last one though :[

Colton
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
May I please ask for the step-by-step of one more? O:] Thank you very much, I would greatly appreciate it... I apologize... I'm at the end of the chapter almost and they're getting progressively more difficult for me...

\(\displaystyle

\frac {\frac{1}{a^2} + \frac{2}{ab} + \frac{1}{b^2}}{\frac{1}{a^2} - \frac{1}{b^2}}

\)

Once again, sorry for posting more than one.... this one's got me just as stumped as the last one though :[

Colton
\(\displaystyle

\frac {\frac{1}{a^2} + \frac{2}{ab} + \frac{1}{b^2}}{\frac{1}{a^2} - \frac{1}{b^2}}

\)

\(\displaystyle = \frac{\frac{b^2}{a^2b^2} + \frac{2ab}{a^2b^2} + \frac{a^2}{a^2b^2}}{\frac{b^2}{a^2b^2} - \frac{a^2}{a^2b^2}}\)

\(\displaystyle = \frac{\frac{b^2 + 2ab + a^2}{a^2b^2}}{\frac{b^2 - a^2}{a^2b^2}}\)

\(\displaystyle = \frac{b^2 + 2ab + a^2}{b^2 - a^2}\)

\(\displaystyle = \frac{(b + a)^2}{(b - a)(b + a)}\)

\(\displaystyle = \frac{b + a}{b - a}\).
 
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Nov 2009
21
0
California, United States
Prove It,

Thank you very, very much for all your help. That provided much assistance. :]

Colton