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March 6th, 2017, 10:59 AM   #11
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Originally Posted by Denis View Post
Nooooooooooo.....
Try same procedure with (10 - 1)/15 ; you'll see why
I do!! Thank you. I am not reducing terms properly.

For my understanding - is the numerator $\displaystyle 2u^2-1$ considered a term in its entirety or can one 'pull out' the variable? I seem to remember (perhaps incorrectly) that you can reduce variables across the division sign by subtracting the powers.

Is there a reference that I can review for this rule? Can someone name the rule or where to find it?

Thanks again!
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March 6th, 2017, 11:00 AM   #12
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The wording "reduce down our u's" is just as vague as "take out our like terms".
Can you tell me the proper name of what its called to reduce common terms? That is what I am trying to do, the question now is whether or not I truly have common terms!

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March 6th, 2017, 11:29 AM   #13
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Originally Posted by chopnhack View Post
For my understanding - is the numerator $\displaystyle 2u^2-1$ considered a term in its entirety or can one 'pull out' the variable? I seem to remember (perhaps incorrectly) that you can reduce variables across the division sign by subtracting the powers.

Is there a reference that I can review for this rule? Can someone name the rule or where to find it?
I suggest:
1: look up "law of exponents"
2: google "simplifying algebraic fractions"
3: get a girlfriend who's a math expert
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March 6th, 2017, 12:15 PM   #14
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Originally Posted by chopnhack View Post
I do!! Thank you. I am not reducing terms properly.

For my understanding - is the numerator $\displaystyle 2u^2-1$ considered a term in its entirety or can one 'pull out' the variable? I seem to remember (perhaps incorrectly) that you can reduce variables across the division sign by subtracting the powers.

Is there a reference that I can review for this rule? Can someone name the rule or where to find it?

Thanks again!
Terms are separated by addittion or subtraction so $2u^2 - 1$ is 2 terms. Something like $u^3v^5wxy^7$ is only 1 term , whereas $x-y+z$ is 3 terms

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March 6th, 2017, 12:30 PM   #15
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Although there are some videos and other study materials available on the internet, it would be a big advantage to use an elementary algebra textbook. It could be an old textbook that's out of copyright and available on the internet or a physical book. It won't matter if it's a bit dated, as the relevant terminology hasn't changed much. One can learn algebra over a period of six years or more at school, so don't expect to cover it all in just a week.
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March 6th, 2017, 12:41 PM   #16
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On the u front:

$\displaystyle \frac{\sqrt{4 - x^2} - \frac{1}{2\sqrt{4 - x^2}}}{4 - x^2} = \frac{u - \frac{1}{2u}}{u^2}$

so this can further be simplified by creating the same denominator for the top half of the nested fraction:

The phrase that you want is "common denominator."

$\displaystyle \frac{u - \frac{1}{2u}}{u^2}$

becomes:

$\displaystyle \frac{\frac{2u*u}{2u}-\frac{1}{2u}}{u^{2}}$ Fine


becomes:

$\displaystyle \frac{\frac{2u*u-1}{2u}}{u^{2}}$

You can simplify $2u * u = 2u^2$

using the property of (a/b)/c is equivalent to a/bc gives:

$\displaystyle \frac{2u*u-1}{2u*u^{2}}$

from here I can see two paths to further resolve the equation, but one makes the equation zero, which I will omit since it doesn't make sense:

$\displaystyle \frac{2u^{2}-1}{2u^{3}}$

I see the ability to reduce down our u's:

You can cancel a common factor in both the numerator AS A WHOLE and the denominator AS A WHOLE, but is u a common factor of the numerator as a whole? Here is where you go off the rails.

$\displaystyle \frac{2(1)-1}{2u}$

or $\displaystyle \frac{1}{2u}$ = $\displaystyle \frac{1}{2\sqrt{4 - x^2}}$

Unfortunately I know that this is not the correct answer. Please assist me in finding the flaw in the process, thanks!
.
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March 6th, 2017, 01:16 PM   #17
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Originally Posted by skipjack View Post
Although there are some videos and other study materials available on the internet, it would be a big advantage to use an elementary algebra textbook. It could be an old textbook that's out of copyright and available on the internet or a physical book. It won't matter if it's a bit dated, as the relevant terminology hasn't changed much. One can learn algebra over a period of six years or more at school, so don't expect to cover it all in just a week.
Do you think I should start with a high school algebra text or a precalculus text? I have gone through high school and college, but it has been some time. Apparently why I am so rusty.... I thought I had a good grip on algebra...

I mention precalc. because it is my understanding that most of those texts cover some basic algebra before advancing.

One week, no... two months, eh... maybe! I really am looking to ferret out the key concepts that I may have forgotten or mislearned.

Thanks again for your advice!
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March 6th, 2017, 01:23 PM   #18
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Originally Posted by JeffM1 View Post
2u2−12u32u2−12u3

I see the ability to reduce down our u's:

You can cancel a common factor in both the numerator AS A WHOLE and the denominator AS A WHOLE, but is u a common factor of the numerator as a whole? Here is where you go off the rails..
That was one of my earlier questions, whether or not 2*u^2-1 was considered one term or two. When you say a whole, do you mean that the entire expression must be in both numerator and denominator?

Thanks in advance.
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March 6th, 2017, 02:39 PM   #19
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$\dfrac{\sqrt{4-x^2}-\frac{1}{2\sqrt{4-x^2}}}{4-x^2} \cdot \dfrac{2\sqrt{4-x^2}}{2\sqrt{4-x^2}}$

$\dfrac{2(4-x^2) - 1}{2(4-x^2)^{3/2}}$

$\dfrac{7-2x^2}{2(4-x^2)^{3/2}}$
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March 6th, 2017, 02:42 PM   #20
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That was one of my earlier questions, whether or not 2*u^2-1 was considered one term or two. When you say a whole, do you mean that the entire expression must be in both numerator and denominator?

Thanks in advance.
Yes

Remember that an expression inside parentheses merely represents a single number. You can view the expression as a single term.

$\dfrac{ac + ad}{ab} = \dfrac{\cancel a(c + d)}{\cancel ab} = \dfrac{c + d}{b}.$

$a,\ b,\ c,\ d \ne 0\ and\ a \ne 1 \implies$

$\dfrac{c + d}{b} = \dfrac{a(c + d)}{ab} = \dfrac{ac + ad}{ab} \ne \dfrac{ac + d}{ab}.$

The problem with "pre-calc" is that there does not seem to be a standard definition for it. In some schools, it is college algebra. In others it is an introduction to analysis and applications of calculus.

Because you previously studied algebra, I'd start with a high school algebra text. It may be that with a bit of practice, it all comes back to you. If, however, it is a struggle, you are not ready for calculus. If you breeze through a high school algebra text getting almost 100% of the problems with answers right, get a college algebra text. In college, they generally cover all the algebra (except maybe for trig functions) you will need for calculus in a single year. See if you get that easily. If not, take a college algebra course before you tackle calculus.
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