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someone solve these, please
icecreamisyummy
post Nov 24 2007, 09:26 AM
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okay so i'm trying to get ready for a math placement test, and well last time i took algebra was...my junior year in high school. so im a little rusty but can someone solve these for me

1. (3x-2y)^2=
A. 9x^2-4y^2
B. 9x^2+4y^2
C. 9x^2-4y^2-6xy
D. 9x^2-4y^2-12xy

2. x^2-x-6/x^2-4=
A. x-3/2
B. x-3/x-2
c. x-3/x+2
D. 3/2

okay so I have the answers to them. This is suppose to be elementary algebra, and I feel dumb that I can't freaken figure out how they got the answers. Yea, those are the only ones I had trouble with. The one I'm having the most trouble with is 2. Idk, I really do feel dumb, so go ahead, say I'm dumb lol.

anyway, I'll tell you guys if youre right or wrong.
and I would like an explanatin on how to do these, specially number 2

I'm sure after I see the explanation, a light bulb with go off lol

oh and srry for posting this on here, I figured I would get more responses and get em a lot faster.

EDIT//

okay so i figured out the first one... and almost figured out the second one so can someone help!! i wanna make sure i figured it out correctly
 
livwho
post Nov 24 2007, 12:01 PM
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2. x2-x-6/x2-4=

Answer: B
 
icecreamisyummy
post Nov 24 2007, 12:14 PM
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QUOTE(ThomasC @ Nov 24 2007, 12:03 PM) *
#1 is B as well.


no number 1 is NOT B. thats what i thought too but it isnt.
 
lojay
post Nov 24 2007, 12:29 PM
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Number one is A
I'm not really sure how you set up number two.. is it x squared minus x minus 6 over x squared minus 4?

If so... then number two would be B.

The key to these is pretty much foiling. [First Outer Inner Last]

For the first problem you foil (3x-2y)(3x-2y) and get 9x^2 - 6xy - 6xy + 4x^2 and that simplifies to 9x^2 + 4y^2


For the second problem you change x^2 - x - 6 to (x-3)(x+2) and change x^2 - 4 to (x+2)(x-2) and now since there is x+2 on top of the divding line and underneath it you can cancel them both out so you are left with (x-3) over (x-2).


Hope I helped...
 
icecreamisyummy
post Nov 24 2007, 12:34 PM
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QUOTE(elbaliava @ Nov 24 2007, 12:29 PM) *
Number one is A
I'm not really sure how you set up number two.. is it x squared minus x minus 6 over x squared minus 4?

If so... then number two would be B.

The key to these is pretty much foiling. [First Outer Inner Last]

For the first problem you foil (3x-2y)(3x-2y) and get 9x^2 - 6xy - 6xy + 4x^2 and that simplifies to 9x^2 + 4y^2
For the second problem you change x^2 - x - 6 to (x-3)(x+2) and change x^2 - 4 to (x+2)(x-2) and now since there is x+2 on top of the divding line and underneath it you can cancel them both out so you are left with (x-3) over (x-2).


Hope I helped...


youre wrong about number one. it is NOT A or B
 
pandamonium
post Nov 24 2007, 12:34 PM
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then number 1 is D. lol i havent taken trig/algebra since high school too so dont feel bad, i did them just to freshen, up i will update when i get number 2 done.
 
lojay
post Nov 24 2007, 12:38 PM
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Oh crap. haha, my bad. Meant 9x^2 - 12xy + 4y^2

In which case, yeah, it is D. Sorry. My math was right on my paper, just read the wrong answer, haha.
 
icecreamisyummy
post Nov 24 2007, 12:38 PM
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QUOTE(pandamonium @ Nov 24 2007, 12:34 PM) *
then number 1 is D. lol i havent taken trig/algebra since high school too so dont feel bad, i did them just to freshen, up i will update when i get number 2 done.

youre right!! about number 1, but can you explain to me how you got -12xy.

my uncle told me you multiple the inside twice or somethig like that... 3x * 2y=6xy * 2= 12xy. somethign along those lines
 
pandamonium
post Nov 24 2007, 12:39 PM
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did i get it right?

YESSSS!!!


number one cause

it turns to be

9x^2-6xy-6xy+4y^2

then

9x^2-12xy+4y^2

cause both of the (-6xy) added together makes (-12xy)

edit............

you can combine x & y as long as they are like terms and on the same side.
 
icecreamisyummy
post Nov 24 2007, 12:41 PM
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QUOTE(pandamonium @ Nov 24 2007, 12:39 PM) *
did i get it right?

number one cause

it turns to be

9x^2-6xy-6xy+4y^2

then

9x^2-12xy+4y^2

cause both of the (-6xy) added together makes (-12xy)


hmm okay
 
lojay
post Nov 24 2007, 12:43 PM
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Here, I'll scan in all my work on paper so that you can see it. I'll just rewrite it first so that it isn't so confusing.

And yes, that was right pandamonium
 
icecreamisyummy
post Nov 24 2007, 12:45 PM
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QUOTE(ThomasC @ Nov 24 2007, 12:43 PM) *
So you got both answers now?

i knew both answers already, i jstu couldnt figure out what iw as doing wrong and how i needed to solve them.
 
pandamonium
post Nov 24 2007, 12:47 PM
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i forgot how to do fractions lMAO so imma wait till you scan the answers in. i feel so behind. good thing im taking advanced algebra next sem.
 
icecreamisyummy
post Nov 24 2007, 12:50 PM
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QUOTE(pandamonium @ Nov 24 2007, 12:47 PM) *
i forgot how to do fractions lMAO so imma wait till you scan the answers in. i feel so behind. good thing im taking advanced algebra next sem.

i have more problems :-( .... but theres are suppose to be "college level mathematics"

i need to remember how to do this sh*t if not im gonna be stuck in intermediate algebra.and i wont be able to take anatomy. grrr
 
lojay
post Nov 24 2007, 12:58 PM
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I hope that tag works... if not I'll edit it...
Edit// Phew. It's a bit big... but oh well. I hope that makes sense.
 
icecreamisyummy
post Nov 24 2007, 01:22 PM
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okay i def understand number one now lol. no doubts anymore.

but number 2.. i mean i understand it, but...well kinda lol

thanks by the way everyone!
 
*jeanna*
post Nov 24 2007, 01:26 PM
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miy heaed hurets
 
*CowerPointyObjects*
post Nov 24 2007, 02:08 PM
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Elementary algebra? HAHAHA I've taken calculus and exempted out of college math and I can't remember how to do this shit.
 
icecreamisyummy
post Nov 24 2007, 02:12 PM
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QUOTE(CowerPointyObjects @ Nov 24 2007, 02:08 PM) *
Elementary algebra? HAHAHA I've taken calculus and exempted out of college math and I can't remember how to do this shit.

idk if to take this as an insult or something else.
 
*CowerPointyObjects*
post Nov 24 2007, 02:17 PM
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Why in the hell would you take it as an insult? It's just recognition of my complete and total inability to do math.
 
Elba
post Nov 24 2007, 02:25 PM
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LMAO. Sorry, but some of you guys suck at math.
 
Spirited Away
post Nov 24 2007, 02:42 PM
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I have a problem with how you typed up number two. If it supposed to be: (x^2-x-6)/(x^2-4)=

Then, I'd work it out like this:

[(x+2)(x-3)]/(x+2)(x-2)=
(x-3)/(x-2)

It is because both the numerator and denominator (x+2)'s cancel each other out.

I haven't touched math in 4 years so I don't know if that's right... but oh well.
 
Gigi
post Nov 24 2007, 03:00 PM
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^ Thanks for clearing it up; that is COMPLETELY different from what she actually wrote.

And yes, I agree with Elba. Sorry if I'm being insensitive, but honestly...I don't understand why some of you can't expand #1.
 
icecreamisyummy
post Nov 24 2007, 03:10 PM
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QUOTE(Spirited Away @ Nov 24 2007, 02:42 PM) *
I have a problem with how you typed up number two. If it supposed to be: (x^2-x-6)/(x^2-4)=

Then, I'd work it out like this:

[(x+2)(x-3)]/(x+2)(x-2)=
(x-3)/(x-2)

It is because both the numerator and denominator (x+2)'s cancel each other out.

I haven't touched math in 4 years so I don't know if that's right... but oh well.

I copied it exactly from the paper...
 
*CowerPointyObjects*
post Nov 24 2007, 03:16 PM
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To be fair, I just assumed I couldn't do it because I didn't want to think about it. I knew I could do it if I wrote it out, and I did remember how to do it as I was typing something a second ago, but I don't care enough to. I am quite terrible at math, though.
 

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