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Who's willing to help me with a simple algebra pro

 
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liesnalibis View Drop Down
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Post Options Post Options   Thanks (0) Thanks(0)   Quote liesnalibis Quote  Post ReplyReply Direct Link To This Post Posted: Dec 11 2013 at 11:13pm
Originally posted by _ConcreteRose_ _ConcreteRose_ wrote:

Ok let me give you a problem
5 x 3
     ---
      5

How would you solve this?

5      3       1      3       3      
--- x --- = ---- x ---- = --- = 3
1       5      1      1        1

      OR

 5     3      15
--- x --- = ---- = 3
 1     5       5 
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Post Options Post Options   Thanks (0) Thanks(0)   Quote _ConcreteRose_ Quote  Post ReplyReply Direct Link To This Post Posted: Dec 11 2013 at 11:14pm
edit. nvm 

Edited by _ConcreteRose_ - Dec 11 2013 at 11:15pm
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Post Options Post Options   Thanks (0) Thanks(0)   Quote _ConcreteRose_ Quote  Post ReplyReply Direct Link To This Post Posted: Dec 11 2013 at 11:16pm
Originally posted by liesnalibis liesnalibis wrote:

Originally posted by zolloh zolloh wrote:

it's vz=5(x-y)+wz....not vz=5z(x-y)+wz
it's vz=5(x-y)+wz....not vz=5z(x-y)+wz
it's vz=5(x-y)+wz....not vz=5z(x-y)+wz

LOL


Ohhhhhhh so the 5(x-y) is considered a single expression?
yes if you plan on multiplying something by it.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote liesnalibis Quote  Post ReplyReply Direct Link To This Post Posted: Dec 11 2013 at 11:16pm
Originally posted by zolloh zolloh wrote:

it's vz=5(x-y)+wz....not vz=5z(x-y)+wz
it's vz=5(x-y)+wz....not vz=5z(x-y)+wz
it's vz=5(x-y)+wz....not vz=5z(x-y)+wz

LOL


Ohhhhhhh so the 5(x-y) is considered a single expression?
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Post Options Post Options   Thanks (3) Thanks(3)   Quote zolloh Quote  Post ReplyReply Direct Link To This Post Posted: Dec 11 2013 at 11:16pm
Originally posted by liesnalibis liesnalibis wrote:

Originally posted by zolloh zolloh wrote:

it's vz=5(x-y)+wz....not vz=5z(x-y)+wz
it's vz=5(x-y)+wz....not vz=5z(x-y)+wz
it's vz=5(x-y)+wz....not vz=5z(x-y)+wz

LOL


Ohhhhhhh so the 5(x-y) is considered a single expression?



now i can go to sleep LOL
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Post Options Post Options   Thanks (0) Thanks(0)   Quote liesnalibis Quote  Post ReplyReply Direct Link To This Post Posted: Dec 11 2013 at 11:17pm
Originally posted by _ConcreteRose_ _ConcreteRose_ wrote:

Exactly.
Ok, Now solve this one

(z)( y)
   ---
    z

1(y)
----- =y
  1
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Post Options Post Options   Thanks (0) Thanks(0)   Quote _ConcreteRose_ Quote  Post ReplyReply Direct Link To This Post Posted: Dec 11 2013 at 11:17pm
Originally posted by liesnalibis liesnalibis wrote:

Originally posted by _ConcreteRose_ _ConcreteRose_ wrote:

Exactly.
Ok, Now solve this one

(z)( y)
   ---
    z

1(y)
----- =y
  1
exaclty, now do the same thing with this fraction.
(z) (5(x-y))
     --------
        z

what are you left with?


Edited by _ConcreteRose_ - Dec 11 2013 at 11:18pm
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Post Options Post Options   Thanks (0) Thanks(0)   Quote liesnalibis Quote  Post ReplyReply Direct Link To This Post Posted: Dec 11 2013 at 11:17pm
Originally posted by _ConcreteRose_ _ConcreteRose_ wrote:

edit. nvm 

No don't give up!
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Post Options Post Options   Thanks (0) Thanks(0)   Quote liesnalibis Quote  Post ReplyReply Direct Link To This Post Posted: Dec 11 2013 at 11:19pm
Originally posted by _ConcreteRose_ _ConcreteRose_ wrote:

Originally posted by liesnalibis liesnalibis wrote:

Originally posted by zolloh zolloh wrote:

it's vz=5(x-y)+wz....not vz=5z(x-y)+wz
it's vz=5(x-y)+wz....not vz=5z(x-y)+wz
it's vz=5(x-y)+wz....not vz=5z(x-y)+wz

LOL


Ohhhhhhh so the 5(x-y) is considered a single expression?
yes if you plan on multiplying something by it.

Okay, okay, I'll try to let that sink in.
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Post Options Post Options   Thanks (0) Thanks(0)   Quote Sang Froid Quote  Post ReplyReply Direct Link To This Post Posted: Dec 11 2013 at 11:20pm
vz=5(x-y)+wz <------ correct answer

v=5(x-y)+w <---------your answer

What's the different here?
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