# Algebra help

Discussion in 'General Education' started by dclary, Aug 7, 2007.

1. ### dclaryRookie

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Aug 7, 2007

Hello. I hope I am posting in the correct forum. I am a middle aged Mom who has decided to go back to school and all was good until I met, "ALGEBRA"! For the most part, I am catching on but have run into a problem with a recent assignment. I got a 35% on and to say I was not happy is an understatement; I was sure I understood the material. Anyway, below are two problems that I'm having a real hard time with.

This first one is, for a lack of better words, driving me crazy! I keep getting the same answer.

I tried to add a picture of the problem but it won't let me. It is a fraction

a to the 2nd. b to the 2nd.
-------------------------
a to the 6th.

I keep getting "a" to the 2nd "b" to the 2nd. What am I doing wrong?

The next is more complex and honestly, I am clueless. My 20 year old niece, who is attending college couldn't even help me. It is also in fraction form.....

y with a -3 exp. over "m" then inside of brackets to the right is "m" with an exp. of 4 over "y" with exp. of 2 (then a + sign) "y" with an exp. 3 over "m" with an exp. of -3

It is really hard to type out so hopefully you will understand. I'm not looking for just the answer, I need to understand how to solve them so that I can finish the remaining problems.

Any help would be greatly appreciated.

3. ### AliceaccMultitudinous

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Aug 7, 2007

a to the 2nd. b to the 2nd.
-------------------------
a to the 6th.

When you're dividing, you subtract exponents. So, assuming that only the b is to the second on the top, you get:
a (to the power of 2-6) times b (to the second)
that's a to the -4 b squared, or it could be written as b squared over a to the 4th. (Negative exponents mean take the reciprocal-- that puts the a to the 4th in the bottom!)

I'm having trouble reading the 2nd problem-- do you want to type it in word them PM it to me? Or email it?

4. ### dclaryRookie

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Aug 7, 2007

Thank you so much. I have jpg of the problem but when I tried to add it to the post it wouldn't allow me to. I can email it to you if you don't mind. Thanks again.

5. ### AliceaccMultitudinous

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Aug 7, 2007

OK.

I'm going to bed in a minute, but I'm usually online early in the morning

6. ### dclaryRookie

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Okay, I understand. If you would give me your email addy I can email you the problem. I tried to PM it as well as adding the image in my post but it keeps telling me that I am not allowed to do that. Thanks again for all of your help.

Last edited: Aug 7, 2007
7. ### AliceaccMultitudinous

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Aug 7, 2007

Let's just go with it as you've typed it:

When you divide, you subtract exponents. So that first fraction would become

m to the 3rd y to the -5

added to y to the 3rd over m to the -3.

Or: y with a -3 exp. over "m" then inside of brackets to the right is "m" with an exp. of 4 over "y" w(negative exponent: take reciprocal) plus m to the 3rd y to the 3rd

To add fractions you need LCM. In this case the only denominator is y to the 5th. First fraction doesn't change. Second fraction, multiply both top and bottom by y to the fifth.

so we have: m to the 3rd over y to the fifth
plus
m to the 3rd y to the 8th over y to the fifth

There are no like terms to combine; it could be factored but there's no point.

Final answer: (m to the 3rd plus m to the 3rd y to the 8th) all over y to the 5th.

8. ### dclaryRookie

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Aug 7, 2007

I think I understand the general concept, however, my original answer was wrong which leaves me with two choices and which neither match what you have. The choices are...

m to the 4th over y to the 2nd plus "my" to the 4th. over m to the third

m to the 3rd. over y to the 5th. plus m to the 2nd.

9. ### TeacherGroupieModerator

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Aug 7, 2007

I don't know that I can help much with the math, but let me at least lend aid with the typography.

You can type "m to the 4th" as m^4. This is an old convention from the early days of computers.

The problem as you typed it is
I'm thinking that's

y^-3 . m^4 + y^3
m^01 . y^2 + m^-3

Am I close?

10. ### dclaryRookie

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Aug 7, 2007

Yes, that is it two listed on the right are in brackets

11. ### TeacherGroupieModerator

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Oho, that would make a difference, along the lines of PEMDAS:

(a) The way I typed it, it's the first fraction multiplied by the second and to that lot is added the third.

(b) But it sounds like you intend the first fraction multiplied by the sum of the second and third - is that correct?

12. ### TeacherGroupieModerator

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Aug 7, 2007

There's no non-clumsy way to type that, but if you'll indulge me by pretending that parentheses stacked on top of each other make one big parenthesis, let me try this:

y^-3 . (m^4 + y^3 )
m^01 .(y^2 + m^-3)

Is that it?

13. ### dclaryRookie

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Aug 7, 2007

I think so but honestly, I'm not sure. I am putting the link below to the picture of the problem, however, I can't put it like a typical link so you will have to type in the colon, period (dot), and back slashes to your browser.

http i3 (dot) photobucket (dot) com/albums/y63/blzrcrzy/Image7.jpg

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15. ### TeacherGroupieModerator

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Aug 8, 2007

First, and for the rest of the audience: there are three multiple choice answers given.

My math guru tells me that the appropriate approach is to use the distributive property. Doing so gives

m^4y^-3 + y^-3y^3
my^2^01 + mm^-3

which is answer A on your jpg - but that's a stupid answer as a final answer, because it contains unreduced fractions, O wickedness! Answer B, which contains

my^4
m^3

is stupid for the same reason.

That leaves answer C, which is

m^3 + m^2
y^5

Not that I could get there unaided either. Quoth my math guru:

To get there, simplify answer A. Now y^3y^-3 = 1, and mm^-3 = m^-2, so the second fraction simplifies to

__1__ = m^2
m^-2

And in the first fraction the m in the denominator cancels out one m in the numerator to yield m^3, and we can punt that y^-3 in the numerator down to the denominator in the form of y^3, so the first fraction ends up as

m^3
y^5

16. ### AliceaccMultitudinous

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Aug 8, 2007

Thanks TG. For some reason I just couldn't visualize the problem.

17. ### TeacherGroupieModerator

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Aug 8, 2007

Neither could I, and I'm not you, so I couldn't make any assumptions except that there might possibly be a Cute Trick.

I don't know whether I'm authorized to post attachments on this forum, either.

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19. ### dclaryRookie

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I want to thank the both of you very much. I'm not going to say that I fully understand yet, however, I am going to take the information you gave and try to complete some of the others I got wrong.

Just to get it clear in my head, I have one more question. When I have a problem such as the one below, am I just subtracting? For example:

y^7
----
y^9

y= -2

I want to thank you again, there is so much information when dealing with Algebra and it is all so confusing.

20. ### teachmemathCompanion

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Aug 8, 2007

yes when you divide you subtract the exponents this answer would be y^2

21. ### dclaryRookie

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Aug 8, 2007

It is y^-2 correct? My original answer was y^2 and it was wrong.

22. ### teachmemathCompanion

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Aug 8, 2007

Yes that's right. I'm sorry I forgot to put the negative

23. ### dclaryRookie

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No problem, I just wanted to double check. Boy, this stuff is hard! When I was in school, Algebra was an elective and not required. Of course, most of us girls took what they called, "commericial math" and believe me, it was nothing like this....lol

24. ### teachmemathCompanion

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Aug 8, 2007

lol. If you have anymore questions in the near future,don't hesistate to ask us on the board for help
it just takes practice and a good teacher

25. ### dclaryRookie

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Aug 8, 2007

Thank you again to everyone for the help, it is much appreciated. You guys have a great "rest of the week"!

Take care,
Diana

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Aug 8, 2007

And remember, that y^-2 is equal to 1/y^2. Or, if you see a negative in the numerator of a fraction, move the coefficient to the denominator but use a positive exponent.

example: y^-2 x^3 divided by y^4 x^-1 would be rewritten as

y^4 x^3 x^1 divided by y^4 y^2 which would equal

y^4 x^4 divided by y^6 after you combine the common terms .

Also remember that in order to add the exponents for multiplication, or subtract them for division, the terms need to have the same bases.

27. ### dclaryRookie

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Aug 8, 2007

Thank you Upsadaisy. This is all so confusing...lol

28. ### TeacherSandraEnthusiast

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Aug 8, 2007

I have to say that I have always hated Math because I did so terribly; I didn't have a clue! I am so thankful that both my kids do so well with Math/Algebra.

This brochure came to my box the other day and I am saving it just in case my soon to be 7th grader begins to have trouble; so far his teacher says he's tops in the class.

You might be interested in this DVD (no, I am not a distributor, etc..)
just want to pass along what I received.

THE GREAT COURSES by The Teaching Company; there's also a website:
www.theteachingcompany.com

Algebra 1 - Profession Monica Neagoy; Course #101 - 30 lesons (30 minutes/lesson) DVD on sale for \$79.95

There is a list of daily topics/equations, etc...

Good luck,
Sandra

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