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New Member
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Feb 4, 2009, 03:25 PM
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simplify algebra
Well I am struggling a bit with this I hope I'm on the right lines but some how I get this nagging feeling I'm all at sea. I have to simplify these expressions am I going about them right.
3x-1+4*5= 3x+4x-1+5 = 7x+4
2(t+6)-3(4-3t) = 2*t +2*6-3*4+3*3t = 2t +12-12 +9t=2t +9t +12 -12 =11t
12-2p/3 - p/3 = 12/3- p/3 = 4-p/3
Have tried to give all my steps and fingers crossed I'm not too far away, any help most welcome
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Uber Member
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Feb 4, 2009, 03:39 PM
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I can see the struggles:
#1: What happened to the 4*5; isn't 4*5 20
#2: Try not to do the sign change right away. Take -3(4-3t) as
(the first expression) - (the second expression of 3(4-3t) so you get -(12+9t); now do the - stuff and get -12-9t. It's an extra step, but the sign changes will cause problems.
#3: Look at it as -2/3 of p - 1/3 of p or or -p
Think of this as a number line. Your at -2/3 and then you go negative again 1/3. Where do you end up?
Want to try again?
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New Member
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Feb 4, 2009, 04:37 PM
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OK will try last one first
#3 I think I see what your saying if I was at -2/3p and progressed by removed (subtracting ) p/3 id be at - p/3 so would that then be 4 ( because 12/3 = 4 )
making the answer 4-p/3
#2 Im not quite up to speed on this one sorry I understand what your saying, take the second part -3(4-3t) which gives me -(12+9t) but not sure what happensafter that sorry
#1 yeah would help if I copied it out right but yes I see the 20
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Uber Member
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Feb 4, 2009, 05:21 PM
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#3: 12-p
Subtracting -1 from say -2 then you get -3. If you subtracted 1 from -2 you would get -1
#2: -12+9t; so add that to the first expression
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New Member
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Feb 5, 2009, 08:33 AM
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Right new day new thinking question 3
It is written like this
12-2p/3 - p/3 so I could rewrite it as
12-2p-p / 3 which would give me 12 -3p/3
Or 4-p answer
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Uber Member
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Feb 5, 2009, 08:51 AM
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No: it's 2/3p-1/3p or (-2/3-1/3)p or (-2-1)/3 or -3/3p
-3/3 is -1, so the answer is
12-1p or 12-p
You can't mix apples and oranges. P's and numbers don't mix initially.
You can, however, factor out common factors.
e.g. 12x-12 = 12(x-1) or 12x-3 I= 3(4x-1)
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New Member
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Feb 5, 2009, 10:15 AM
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Gee at this rate ill need 6 months to get my head round this
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New Member
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Feb 5, 2009, 10:50 AM
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I really do thank you for your endeavor keepitsimplestupid. I just don't seem to be able to look at algebra and see what I should do.
I don't quite understand stating the obvious I know but if the whole thing is divided by 3 why doesn't the 12 end up as 12/3 = 4
perhaps if I try to explain what's going on in my head you might be able to shake some sense into it. Right to start with:
12-2p/3 -p/3 = well well I thought if I split it to read as follows 12 /3 and -2p combine the other -p/3 so would end up starting the p part first -2p-p =-3p so -3p/p = -1 or rather -1p (-p) and 12/3 =4 so came up with 4 -p . I think if the 12 was not on top line I wouldn't have divided it by three, but because they have it there thought you should.
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Uber Member
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Feb 5, 2009, 10:54 AM
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I actually want to thank you for the way in which you submitted your problems.
You submitted both the problem and attempted work. This is very good. We also went iterations, which is good too.
Remember that you can check your answers by substituting a number like 1 and compare the results of the original problem and the simplification. Remember, that using 1 may not be the easiest number to use.
Thanks.
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Uber Member
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Feb 5, 2009, 10:59 AM
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The problem is this: 12-2p/3 -p/3 not this (12-2/3p-1/3p)/3
Take my anser of 12-1p and substitute 1
You get 12-(1)(1) or 11
Take the original problem of 12-2p/3 -p/3 :
12-(2)(1)/3-1/3
12-2/3-1/3
12-3/3
12-1
11
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Uber Member
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Feb 5, 2009, 11:27 AM
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annother way of looking at is is:
Collect the number of p's and collect the number of numbers.
The p's
-1/3
-2/3
The numbers
12
Add the number of p's -1/3+-2/3 = -3/3 = -1
Add all of the numbers
12
So you have -1p+12 or 12-1p
That will work all the time and it will allow you to get faster. Right now though, I'll bet they want you to show your work.
At this level, you just collecting like terms.
You can have -3x-6y+7z+20-11-1-1/3y
Doesn't matter:
#'s: 20, -11, -1, (20+-11+-1)
x's: -3 --> 3
y's: 6, -1/3 --> 6 minus 1/3
z's: 7 --> have 7
Then put them all together
You don't have to play the game of moving all the stuff around.
Basically the sign will screw up more problems than you can image. What I just showed, I think, is a different approach. Notice that it's easier.
Also, I will emphasize that you always use a script x and a script y. Because with x's and y's it's sometimes easy to mistake them.
Penmanship and neatness is extremely important.
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Uber Member
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Feb 5, 2009, 01:09 PM
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Hopefully the post will get merged.
Even this thing, doing it my new way:
2(t+6)-3(4-3t)
Look at the problem this way: 2(t+6)+(-3)(4-3t)
Count the t's
2, 9
Count the numbers
12, -12 of them
Answer 11t
That's the most efficient way of doing the problem.
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New Member
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Feb 8, 2009, 02:03 PM
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Well Im hoping I finally managed to get my head around what you been saying. I think the idea of the extra step with the double - is a good one and makes lot of sense. Now here's a thought on the last question #3 (could be foot on the mouth time) well if you manipulate it as follows would this be realistic?
12-2p/3 - p/3
Multiply both parts by 3 gives you
(36-6p) - 3 p
Divide by 3
12-2p - p
Then minus the end p from 2p gives p
Giving answer 12 - p
Or have I just managed to mangle a n answer to fit the correct one.
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Uber Member
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Feb 8, 2009, 04:42 PM
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Try your multiply by 3 again. 36-2p-1p---> 36-3p--->12-p
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New Member
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Feb 9, 2009, 12:07 PM
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Well I had feeling I had rumbled that one but am I right in thinking that you combined the -2p and the -1p thus 36-3p then divide by 3.
Is it realistic to think by combining the like letters as early as possible it tends to make it easier as you progress with algebra ( like I really think I will ever progress)
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Uber Member
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Feb 9, 2009, 12:21 PM
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That's generally called collecting terms.
As I was trying to say: You can do this somewhat by inspection, if you do everything somewhat "at once". As your scanning the expression, do the simple multiplys and apply them to the terms.
If you don't, things will end up MESSY fast.
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