Adding two rational expressions?

[Oilaye]

Hello, I have a question regarding simplification of rational expressions. I understand two rational expressions need a like denominator (or, LCD) in order to add/subtract them, the same as any two fractions. The issue comes in when I take into account opposite factors. I believe it would be best to show you my dilemma and ask from there.

Here is the picture of what I'm about to ask: http://puu.sh/3BylW.png

I understand you can factor a -1 from either of the expressions opposing factors. It doesn't matter which you factor from because you'll get the same answer in the very end, after fully factoring everything.

The issue is, when asked to write the LCD, we include the -1 as part of it. But when operating on these, we include the negative 1 as a factor only to the expression we've factored it from (well, from the expression with the opposite factor we've factored it from). This has confused me. I thought the LCD should be the same for each rational expression. If I set the LCD with the -1 for both of the denominators, then I get something different and end up messing it up. It doesn't feel right to say the LCD should include the -1 for both if we're only using it on one of the expressions. So then how is this correct?

And just to make it convenient, when I included the -1 as part of the LCD for both of the expressions I end up with:

1/-(x - 4)(x + 2) + 14x/-(x - 4)(x + 2)

-1/(4 - x)(x + 2) - 14x/(4 - x)(x + 2), which is also (-1 - 14x)/(4 - x)(x + 2)

I'm not sure what happened, but this is not the same answer I would have gotten had I used the -1 on only one of the expressions. I am lost here.

I apologize in advance if something isn't clear. I'm still trying to learn how to properly ask questions in math. I will be glad to clear up anything that isn't clear!

Thank you for your help, I do appreciate it.

 

[MarkFL]

If you are told to find the LCD of the two expressions:

\(\displaystyle \dfrac{1}{a},\,\dfrac{1}{-b}\)

then indeed, you should give \(\displaystyle -ab\) as the answer. However, if you are told to add the expression, then I would write:

\(\displaystyle \dfrac{1}{a}+\dfrac{1}{-b}=\dfrac{1}{a}-\dfrac{1}{b}\)

and now I would use an LCD of \(\displaystyle ab\)

You will get the same result either way:

\(\displaystyle \dfrac{1}{a}+\dfrac{1}{-b}=\dfrac{1}{a}\cdot\dfrac{-b}{-b}+\dfrac{1}{-b}\cdot\dfrac{a}{a}=\dfrac{-b}{-ab}+\dfrac{a}{-ab}=\dfrac{-b+a}{-ab}=\dfrac{b-a}{ab}\)

\(\displaystyle \dfrac{1}{a}-\dfrac{1}{b}=\dfrac{1}{a}\cdot\dfrac{b}{b}-\dfrac{1}{b}\cdot\dfrac{a}{a}=\dfrac{b}{ab}-\dfrac{a}{ab}=\dfrac{b-a}{ab}\)

 

[Oilaye]

If you are told to find the LCD of the two expressions:

\(\displaystyle \dfrac{1}{a},\,\dfrac{1}{-b}\)

then indeed, you should give \(\displaystyle -ab\) as the answer. However, if you are told to add the expression, then I would write:

\(\displaystyle \dfrac{1}{a}+\dfrac{1}{-b}=\dfrac{1}{a}-\dfrac{1}{b}\)

and now I would use an LCD of \(\displaystyle ab\)

You will get the same result either way:

\(\displaystyle \dfrac{1}{a}+\dfrac{1}{-b}=\dfrac{1}{a}\cdot\dfrac{-b}{-b}+\dfrac{1}{-b}\cdot\dfrac{a}{a}=\dfrac{-b}{-ab}+\dfrac{a}{-ab}=\dfrac{-b+a}{-ab}=\dfrac{b-a}{ab}\)

\(\displaystyle \dfrac{1}{a}-\dfrac{1}{b}=\dfrac{1}{a}\cdot\dfrac{b}{b}-\dfrac{1}{b}\cdot\dfrac{a}{a}=\dfrac{b}{ab}-\dfrac{a}{ab}=\dfrac{b-a}{ab}\)

Thank you for your response!

What you said makes absolute sense to me. But for some reason when the problem gets larger (like the one I linked,) I get thrown off. I understand it to be using the same principle, but I believe I'm overlooking something somewhere.


Please note that these aren't very long problems. I'm only wanting to understand what I'm not seeing. I know I'm missing something but I...am not getting it. I've spent a couple hours on this and just don't see what I'm doing wrong. You mention it working either way, but it isn't for me. I can't use -1(x-4)(x + 2) for the LCD and get the same answer. It only works when I apply the -1 to the expression I've factored the -1 from (well, it's opposite factor anyway). >.<

I appreciate your response Mark. I am very grateful for you taking the time to answer me. ^^

 

[lookagain]

1/[-(x - 4)(x + 2)] + 14x/[-(x - 4)(x + 2)]

-1/[(4 - x)(x + 2)] - 14x/[(4 - x)(x + 2)], which is also (-1 - 14x)/[(4 - x)(x + 2)]

\(\displaystyle \ \ \ \ \ \)Oilaye, you must have grouping symbols around those denominators, as I inserted with bold brackets above for example, because of the Order of Operations.

 

[JeffM]

The images you have posted are not legible for me.

Someone will be happy to review your work if you will copy it out on this site.

 

[Oilaye]

The images you have posted are not legible for me.

Someone will be happy to review your work if you will copy it out on this site.

Hello Jeff. I got some help with my issue yesterday from a classmate. I understand what's been tripping me up now. I really over-complicated things. Sticking to what a previous poster said about using the -1 with only the expression I factored it from is the best way. I was trying to ask if using -1(x - 4)(x + 2) for the denominator in both of the expressions would yield the same result had I used the -1/-1 * expression negative is being factored from. But I only get confused when I try that. I will stick to the multiplication by -1/-1 way.

That said, I do have one very minor question still remaining. If you have say:

You can factor a -1 from the (x - 4) in the bottom factor. When you do, you would apply a -1 to your numerator. My question is, once you do this, does that -1 affect (-14x - 1), or just the -14x? And before that even, does the - that is already in front of the 14x affect the -1 behind it? If I were to factor the -1 from the numerator, would I have -1(14x + 1)?

And yes, this is is my first Latex equation. Pretty neat.

 

[Deleted member 4993]

Hello Jeff. I got some help with my issue yesterday from a classmate. I understand what's been tripping me up now. I really over-complicated things. Sticking to what a previous poster said about using the -1 with only the expression I factored it from is the best way. I was trying to ask if using -1(x - 4)(x + 2) for the denominator in both of the expressions would yield the same result had I used the -1/-1 * expression negative is being factored from. But I only get confused when I try that. I will stick to the multiplication by -1/-1 way.

That said, I do have one very minor question still remaining. If you have say:

....... This translates to - following Order of Arithmetic Operations (PEMDAS) →

\(\displaystyle \displaystyle -14x \ - \ \frac{1}{(x-4)} (x+2)\)

If you wanted to express:

\(\displaystyle \displaystyle \frac{-14x - 1}{(x-4) (x+2)}\)

you should have written in ASCII format as:

(-14x - 1)/[(x-4)(x+2)]

Notice those parentheses - those are very important to express order of operation

You can factor a -1 from the (x - 4) in the bottom factor. When you do, you would apply a -1 to your numerator. My question is, once you do this, does that -1 affect (-14x - 1), or just the -14x? And before that even, does the - that is already in front of the 14x affect the -1 behind it? If I were to factor the -1 from the numerator, would I have -1(14x + 1)?

Answer to your question depends on the proper expression.

Your questions worry me - are you trying to teach yourself algebra - online?

If yes - don't! - get a face-to-face tutor. There are many retired people (like me) who would love to teach a student like you face-to-face - free - you just have to find them.

And yes, this is is my first Latex equation. Pretty neat.

.

 

[Oilaye]

Hello Khan!

I tried learning online before on my own. That didn't go so well. Right now I'm in a traditional classroom setting taking the Algebra course. I don't have questions all the time, but when I do they are extremely bothersome to me, like this one in particular. I practiced for a few hours on other similar problems. I believe I'm fine with this concept now. But man did it throw me for a loop at first.

As far as getting a tutor, I've always dreamed of having one. It might sound silly to say I've always dreamed of having one, but it's true. I was too concentrated on helping with bills growing up. I did not grow up in the best of environments. Back then I thought learning math was impossible for the common person (lol, right? xD). But now I see that it isn't. You just need to be dedicated. And I am now. I would love to get a tutor but all I've seen so far require money. I never even thought there may be people out there willing to do it for free. I will look more into that.

Appreciate the help everyone. And just for the the record, yes I did mean (-14x - 1)/[(x-4)(x+2)].

 

[JeffM]

Hello Khan!

I tried learning online before on my own. That didn't go so well. Right now I'm in a traditional classroom setting taking the Algebra course. I don't have questions all the time, but when I do they are extremely bothersome to me, like this one in particular. I practiced for a few hours on other similar problems. I believe I'm fine with this concept now. But man did it throw me for a loop at first.

As far as getting a tutor, I've always dreamed of having one. It might sound silly to say I've always dreamed of having one, but it's true. I was too concentrated on helping with bills growing up. I did not grow up in the best of environments. Back then I thought learning math was impossible for the common person (lol, right? xD). But now I see that it isn't. You just need to be dedicated. And I am now. I would love to get a tutor but all I've seen so far require money. I never even thought there may be people out there willing to do it for free. I will look more into that.

Appreciate the help everyone. And just for the the record, yes I did mean (-14x - 1)/[(x-4)(x+2)].

There is a huge difference from the tutor's perspective on who is being tutored. When I was in graduate school, I made some extra money being a tutor in the computer lab. For 90% of those students, the only thing in the world that made me put up with them was that I was being paid. There was also a teacher I liked, who asked me to tutor other grad students in how to read certain kinds of texts. We did our work in the grad student bar, and I was happy to do the work for a few free beers. The difference was in the attitude of the people being tutored. Many people are happy to teach people who are willing to work at learning.

 

[HallsofIvy]

There is a huge difference from the tutor's perspective on who is being tutored. When I was in graduate school, I made some extra money being a tutor in the computer lab. For 90% of those students, the only thing in the world that made me put up with them was that I was being paid. There was also a teacher I liked, who asked me to tutor other grad students in how to read certain kinds of texts. We did our work in the grad student bar, and I was happy to do the work for a few free beers. The difference was in the attitude of the people being tutored. Many people are happy to teach people who are willing to work at learning.

Plus tutoring is much easier after a few beers!

 

[Oilaye]

There is a huge difference from the tutor's perspective on who is being tutored. When I was in graduate school, I made some extra money being a tutor in the computer lab. For 90% of those students, the only thing in the world that made me put up with them was that I was being paid. There was also a teacher I liked, who asked me to tutor other grad students in how to read certain kinds of texts. We did our work in the grad student bar, and I was happy to do the work for a few free beers. The difference was in the attitude of the people being tutored. Many people are happy to teach people who are willing to work at learning.

Thanks for sharing that. I definitely have an interest in learning to understand now. Up through high school I saw math and the more difficult subjects as things I should rush through and not really care about. Wrong crowd. But now I understand. I have come to appreciate math. I enjoy learning new ways it can be applied in everyday situations. It's become a hobby of mine to talk with (ahem, annoy) my math teacher with arbitrary questions regarding how I could use a concept in everyday life. x)