How to Understand a Number Over an Expression

Jason76

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Let's say I was doing this problem:

Simplify:

\(\displaystyle 5(\dfrac{4}{4x - 8}) - 2 = \) - Can the fraction in this expression be viewed as two fractions (as it would if that fraction was flipped)?
 
Let's say I was doing this problem:

Simplify:

\(\displaystyle 5(\dfrac{4}{4x - 8}) - 2 = \) - Can the fraction in this expression be viewed as two fractions (as it would if that fraction was flipped)?

No..
 
Let's say I was doing this problem:

Simplify:

\(\displaystyle 5(\dfrac{4}{4x - 8}) - 2 = \) - Can the fraction in this expression be viewed as two fractions (as it would if that fraction was flipped)?

\(\displaystyle 5(\dfrac{4}{4x - 8}) - 2 = \)

\(\displaystyle (\dfrac{5}{x - 2}) - 2 = \)

\(\displaystyle \dfrac{5 - 2(x-2)}{x - 2} = \)

and continue.....
 
Quit complicating the issue; start like this:
4 / (4x - 8) = 4 / (4(x - 2)) = 1 / (x - 2)

But that's seeing the one fraction as two fractions which Khan said was wrong.

But you guys are seeing the fraction as two fractions:

\(\displaystyle 4(\dfrac{4}{4x } - \dfrac{4}{8}) - 2\)

\(\displaystyle (\dfrac{5}{x - 2}) - 2 = \)

\(\displaystyle \dfrac{5 - 2(x-2)}{x - 2} = \)
 
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I am confused as to what you mean by "as two fractions". I suspect that Subhotosh Khan thought you meant writing \(\displaystyle \frac{4}{4x- 8}\) as \(\displaystyle \frac{4}{4x}- \frac{4}{8}\) . That you cannot do. \(\displaystyle \frac{a}{b+ c}\) is not the same as \(\displaystyle \frac{a}{b}+ \frac{a}{c}\). (Of course, \(\displaystyle \frac{b+ c}{a}\) is \(\displaystyle \frac{b}{a}+ \frac{c}{a}\).)
 
But that's seeing the one fraction as two fractions which Khan said was wrong.

But you guys are seeing the fraction as two fractions:

\(\displaystyle 4(\dfrac{4}{4x } - \dfrac{4}{8}) - 2\) → This is not your original problem

\(\displaystyle (\dfrac{5}{x - 2}) - 2 = \) → This does not follow from above

\(\displaystyle \dfrac{5 - 2(x-2)}{x - 2} = \)

Jason, you are having problems with simple algebra - however, you were trying to do calculus in your previous post. This is recipe for impending disaster.
 
\(\displaystyle (\dfrac{4}{4x } - \dfrac{4}{8})\)

Perhaps the denominator could be made the same, and then it could be seen as two fractions.

Jason, you are having problems with simple algebra - however, you were trying to do calculus in your previous post. This is recipe for impending disaster.

I'm only in Pre-Calculus I over the summer. Plenty of time to fix algebra errors before tackling Calculus.
 
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\(\displaystyle (\dfrac{4}{4x } - \dfrac{4}{8})\)

Perhaps the denominator could be made the same, and then it could be seen as two fractions.
.

Why do you want to make it into two fractions? In the given type of problem, that will only complicate the issue.

Certainly you can write:

\(\displaystyle \displaystyle\frac{1}{ax+b} \ = \ \frac{m}{ax+b} \ + \ \frac{1-m}{ax+b}\)

But that does not progress towards any solution or simplification.
 
\(\displaystyle 5(\dfrac{4}{4x - 8}) - 2 = \)

\(\displaystyle (\dfrac{5}{x - 2}) - 2 = \)

\(\displaystyle \dfrac{5 - 2(x-2)}{x - 2} = \)

and continue.....

We'll just leave it that "We divide \(\displaystyle 4\) by \(\displaystyle 4x - 8\)". Then later we subtract two fractions: \(\displaystyle (\dfrac{5}{x - 2}) - 2 = \) to get the final simplification.
 
We'll just leave it that "We divide \(\displaystyle 4\) by \(\displaystyle 4x - 8\)". Then later we subtract ...

No, that is not what we do.

Seven months ago, we tried to advise you that your learning-approach is backwards.

You replied that your new plan was to review beginning algebra, and then move forward through intermediate algebra, trig, and precalculus. You posted last November that your "main focus is to review algebra..."

That is a great idea, but when are you going to start?

You need to understand the concept of factoring before taking trigonometry or precalculus. Otherwise, you risk missing the big pictures because too much of your concentration will have to be spent on prerequisite matters.

Denis showed you how to cancel factors of 4, in your exercise. After that simplification, we use common denominators to combine algebraic fractions (Subhotosh showed that part).

:idea: Study beginning algebra first; avoid spending so much effort on exercises like this one.
 
Let's say I was doing this problem:

Simplify:

\(\displaystyle 5(\dfrac{4}{4x - 8}) - 2 = \) - Can the fraction in this expression be viewed as two fractions (as it would if that fraction was flipped)?

If you are more comfortable solving equations than simplifying, sometimes letting y = expression helps

\(\displaystyle y = 5(\dfrac{4}{4x - 8}) - 2 \)
\(\displaystyle y+2 = 5(\dfrac{4}{4x - 8}) \)

\(\displaystyle (y+2)(4x - 8) = 20 \)

etc.
 
I'm only in Pre-Calculus I over the summer. Plenty of time to fix algebra <--- False. \(\displaystyle \ \ \ \ \ \ \) > > errors < < <--- That misrepresents your status, Jason76. You have problems with fundamentals of elementary level algebra. \(\displaystyle \ \ \ \ \ \ \)before tackling Calculus.
That is putting the cart before the horse. You should be able to get a "good" grade or higher ** on a comprehensive elementary algebra final exam (actual or sample), and not just a passing grade, to do well in intermediate algebra, and then on to college algebra, and then on to pre-calculus.** Not only should the grade on the hypothetical exam be at that level, but the percentage of each subpart of the course on the final exam should be better than "just passing."
 
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Jason

I am sympathetic to your desire to move ahead rapidly, but I must agree with the many comments saying that the only way to progress at all is to be solid in the preliminary subjects.

Math is progressive. Calculus builds on trigonometry, geometry, and elementary algebra, and they all build on arithmetic. If you do not understand algebra and trigonometry (and some geometry) thoroughly, calculus is a bunch of arbitrary rules, and solving word problems is impossible because the principles involved are all mixed up with algebra, geometry, and trig.

I do not agree, however, that you necessarily need to take umpteen courses. You seem willing to work very hard on your own. What you absolutely need to do is to determine what you do not know. Get a beginning algebra text out of the library and do the exercises that have answers supplied, including the word problems. (I like Khan Academy, but I think that it concentrates too heavily on mechanics at the expense of applying math to problems. That is what word problems are for.)

If you know beginning algebra well, you will race through such a text. Furthermore, make sure you understand the logic behind the procedures. It is not enough to know how to do something. You need to know why to do that something and not some other thing. If you find that there is some aspect of beginning algebra that you do not understand (because you cannot get correct answers on the problems that are answered in the answer key), start doing the other exercises and come here to have your work checked. Given your willingness to work and your apparent intelligence, I suspect that you can review and fully master a first year high school course in elementary algebra in a few weeks. Then do the same with the second year of high school algebra. This is usually called intermediate algebra although it is just a continuation of elementary algebra. Again I suspect that you can review and master this topic in a few weeks on your own. Back when I went to school, they had no course called pre-calculus. Certainly the people who discovered and elaborated calculus in the 17th and 18th centuries had nothing but algebra, geometry, and trigonometry to rely on. So I have no idea whether or not a modern pre-calculus course is particularly helpful to someone who has a solid knowledge of elementary algebra and trig and analytic geometry. Probably teachers of calculus have discovered that the average high school has not taught algebra and trig solidly enough to prepare most students for calculus, and pre-calculus is a way for colleges to remediate the defects of the high schools. (This may be unfair. In my high school, I took a course in my junior year that included trig, three-dimensional Euclidean geometry, some beginning set theory, and some beginning number theory. Except for the trig, I do not remember this grab-bag of a course as being particularly pertinent to calculus, which I studied in my senior year. Of course, students may not be the best judge of the logic behind a curriculum.)

In short, I do not think that you personally need to take four or five courses before you can study calculus with a high likelihood of success. I think that because I think you personally can master the preliminary subjects with the help of some good texts, this site, and some of the resources on the web. But you will never understand calculus without having previously mastered algebra, analytic geometry, and trigonometry.
 
Getting ahead of the math classes

Jason,

What JeffM is true and very important.
Also true is what you said about the weakness of watching random uTube Vids is not the answer.

However, there is no limit to how far ahead you can get. If you have a way to know you have not skipped something important!

Kahn Accademy may be the answer for you...
1) Absolutely Free -- funded by grants from Google, MS, and many others.
2) Moves quickly through things you know.
3) Slows down on things you are learning.
4) Lots of video support
5) You can ask detailed questions here on freeMathHelp:)

https://www.khanacademy.org
Get signedup with a password.

bobbmsee is my name on Khan (if you want a coach)
 
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Right, I agree with you guys. It's not enough to fly thru it. It has to be "beat into your brain" thru drilling. No way around it. Hopefully, I can do enough algebra drills (in my own time apart from Pre-Calclulus I class) to solidify the knowledge. Otherwise, passing with a C might be what I do all thru Calculus and onward, but that's not what I want.
 
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