Solving rational equations

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wesley

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how do you figure out the least common denominator of rational equations that have variables in the denominator.
please show more than one example. i have tried various things like factoring but i can not seem to get the concept. please answer asap. thankyou. please write step by step instructions so i can use it for other rational equations.
 
wesley said:
how do you figure out the least common denominator of rational equations that have variables in the denominator.
please show more than one example. i have tried various things like factoring but i can not seem to get the concept. please answer asap. thankyou. please write step by step instructions so i can use it for other rational equations.
Click to expand...

Please post several example problems - that you could not solve (of which you could not find LCD). Then we can help you constructively.

In general, these follow the same rule of numerical fractions ? factorize the poly nomial and find the least common multiple.
 
Forex:
2/x + 3/x^2 + 1/x
how will you find the least common denominator for this?
Sorry, I fixed the mistake. :D
 
wesley said:
Forex:
2/x + 3/x^2 + 1/x
how will you find the common denominator for this?
Click to expand...

The same way I would find common denominator of:

2/5 + 3/25 + 1/5

I factorize the denominators into their prime factors:

5 = 5 * 1......................... in your case x = x*1
25 = 5 * 5 * 1................... in your case x[sup:2lroradl]2[/sup:2lroradl] = x * x * 1

So the common denominator is:

5 * 5 * 1 = 5[sup:2lroradl]2[/sup:2lroradl] .................. in your case x[sup:2lroradl]2[/sup:2lroradl] = x * x * 1 = x[sup:2lroradl]2[/sup:2lroradl]

Then

2/x * x/x + 3/x[sup:2lroradl]2[/sup:2lroradl] + 1/x * x/x

= 2x/x[sup:2lroradl]2[/sup:2lroradl] + 3/x[sup:2lroradl]2[/sup:2lroradl] + x/x[sup:2lroradl]2[/sup:2lroradl]

= (2x + 3 + x)/x[sup:2lroradl]2[/sup:2lroradl]

= (3x + 3)/x[sup:2lroradl]2[/sup:2lroradl]

= 3 * (x + 1)/x[sup:2lroradl]2[/sup:2lroradl]
 
wesley said:
Forex:
2/x + 3/x^2 + 1/x
how will you find the common denominator for this?
Click to expand...

wesley,
your question should be asking about what is the \(\displaystyle least \ \text{common denominator}\) for this.


Common denominators for this are \(\displaystyle x^2, \ x^3, \ x^4, \ x^5, \ and \ so \ on.\)


But the \(\displaystyle least\) common denominator is the smallest product, such that, all of the original
denominators divide into it.
 
He did specify least - in his original post

how do you figure out the least common denominator of rational equations that have variables in the denominator.
Click to expand...
 
Subhotosh Khan said:
He did specify least - in his original post

how do you figure out the least common denominator of rational equations that have variables in the denominator.
Click to expand...
Click to expand...

That was correct up till that point, but then in his second post, he dropped the word "least,"
as did you in your follow-up post to his.

There must be consistency for everyone to call something what it is, and then to continue
referring to it correctly throughout the discussion.

Subhotosh Khan said:
wesley said:
Forex:
2/x + 3/x^2 + 1/x
how will you find the > > > common denominator < < < for this?
Click to expand...

The same way I would find > > > common denominator < < < of:

2/5 + 3/25 + 1/5

I factorize the denominators into their prime factors:

5 = 5 * 1......................... in your case x = x*1
25 = 5 * 5 * 1................... in your case x[sup:3sqsna40]2[/sup:3sqsna40] = x * x * 1

So the common denominator is:

5 * 5 * 1 = 5[sup:3sqsna40]2[/sup:3sqsna40] .................. in your case x[sup:3sqsna40]2[/sup:3sqsna40] = x * x * 1 = x[sup:3sqsna40]2[/sup:3sqsna40]

Then

2/x * x/x + 3/x[sup:3sqsna40]2[/sup:3sqsna40] + 1/x * x/x

= 2x/x[sup:3sqsna40]2[/sup:3sqsna40] + 3/x[sup:3sqsna40]2[/sup:3sqsna40] + x/x[sup:3sqsna40]2[/sup:3sqsna40]

= (2x + 3 + x)/x[sup:3sqsna40]2[/sup:3sqsna40]

= (3x + 3)/x[sup:3sqsna40]2[/sup:3sqsna40]

= 3 * (x + 1)/x[sup:3sqsna40]2[/sup:3sqsna40]
Click to expand...
 
Stating Range of a funciton

[Question]
I do not understand how you would find the domain and range for the
following: 3/x+2?

[Difficulty]
I thought that the range is related to the horizontal asymptote, and
the numerator degree and denominator degree equal each other. so it
should be the coeffiecients of the 3/1. But the text book says that
y can be anything except zero. why?
I am not in advanced mathematics.


[Thoughts]
ok, so i made the denominator equal to zero and figured out that x
can be anything except negative 2. However, how do i do that for the
range? I thought that the range is related to the horizontal
asymptote, and the numerator degree and denominator degree equal
each other. so it should be the coeffiecients of the 3/1, however i
am totally confused because my textbook says that y can not equal
zero, it can be anything except zero. can u please clarify my
concepts.
 
Re: Stating Range of a funciton

wesley said:
[Question]
I do not understand how you would find the domain and range for the
following: 3/x+2?

[Difficulty]
I thought that the range is related to the horizontal asymptote, and
the numerator degree and denominator degree equal each other. so it
should be the coeffiecients of the 3/1. But the text book says that
y can be anything except zero. why?
I am not in advanced mathematics.


[Thoughts]
ok, so i made the denominator equal to zero and figured out that x
can be anything except negative 2. However, how do i do that for the
range? I thought that the range is related to the horizontal
asymptote, and the numerator degree and denominator degree equal
each other. so it should be the coeffiecients of the 3/1, however i
am totally confused because my textbook says that y can not equal
zero, it can be anything except zero. can u please clarify my
concepts.
Click to expand...

Wesley

A couple of thoughts. First, do not pose a new question on an old thread. People will see that there are multiple replies in the thread and assume that the question was ALREADY answered.

Second, this very site (freemathhelp) has an explanation of range and domain. If you google "range domain" you will find the proper page. Why don't you take a look at it and then come back here if you do not understand what has already been provided. I think the explanation is very simple and clear, but you will probably find that toward the end it gets past where you are yet ready to go. That's OK. You need what is at the beginning.

Third, proper grouping of your expressions is very important. In your question, it is not clear whether f(x) = [(3 / x) + 2] or f(x) = [3 / (x + 2)]. The answers to your question are different for the two different functions.

Fourth, I appreciate that you have described both your thoughts and your difficulty, which is GOOD because it shows you are working the issue, but I must admit I do not understand either of your descriptions. You "made the denominator zero." But a denominator must NEVER be zero. Read the site's explanation, then revise your description of the problem, your difficulty, and your thoughts, and we can talk some more. OK?
 
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