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Apr 18th, 2002, 02:55 PM
#1
Thread Starter
Dazed Member
Combining like terms
Ive been having trouble understanding how to combine like terms when multiplication is distributed over addition. If anyone could help it would be great. Two problems that i have banging my head over are the following
4(2x3 + x2) - 5x2
= 8x3 - x2 'books answer
I end up getting 8x3 + 4x2 - 5x2
Here is another one..... 3(5c) - 15c2 + c
-15c2 + 16c 'books answer
Now 3 * 5c would give us 15c.
15c - 15c2 or 15c + (-15c2)
should gives us 0 c1 which boils down to c
so i end up with c + c.
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Apr 18th, 2002, 03:07 PM
#2
Thats Puzzling!
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Apr 18th, 2002, 03:13 PM
#3
Thread Starter
Dazed Member
Im puzzled by what you find puzzling
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Apr 18th, 2002, 03:20 PM
#4
Hyperactive Member
Re: Combining like terms
Originally posted by Dilenger4
Ive been having trouble understanding how to combine like terms when multiplication is distributed over addition. If anyone could help it would be great. Two problems that i have banging my head over are the following
4(2x3 + x2) - 5x2
= 8x3 - x2 'books answer
I end up getting 8x3 + 4x2 - 5x2
Here is another one..... 3(5c) - 15c2 + c
-15c2 + 16c 'books answer
Now 3 * 5c would give us 15c.
15c - 15c2 or 15c + (-15c2)
should gives us 0 c1 which boils down to c
so i end up with c + c.
I don't know whether you are just joking are not..
But anyway..
1) 4(2x^3 + x^2) - 5x^2
= [(4*2)*x^3] + [4*x^2] - [5*x^2]
= 8x^3 + [4-5](x^2)
= 8x^3 + (-1 * x^2)
= 8x^3 - x^2
2) 3(5c) - 15c^2 + c
= 15*c - 15c^2 + 1*c
= [15 + 1]*c - 15c^2
= 16c - 15c^2
= -15c^2 + 16c
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Apr 18th, 2002, 03:35 PM
#5
Thread Starter
Dazed Member
Thanks thinktank2. I kind of understand your break down of the second problem.
2) 3(5c) - 15c^2 + c
= 15*c - 15c^2 + 1*c
= [15 + 1]*c - 15c^2
= 16c - 15c^2
= -15c^2 + 16c
3(5c) - 15c2 + c
-15c2 + 16c
But why would 3(5c) + c = 16c? wouldn't it be 15c2
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Apr 18th, 2002, 04:16 PM
#6
Hyperactive Member
Re: Combining like terms
About the first one:
Originally posted by Dilenger4
4(2x 3 + x 2) - 5x 2
= 8x 3 - x 2 'books answer
I end up getting 8x 3 + 4x 2 - 5x 2
8x3 - x2 = 8x3 + 4x2 - 5x2
For the second one:
3(5c) = 15c
15c+c = 16c
15c(c) = 15c^2
-Show me on the doll where the music touched you.
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Apr 18th, 2002, 04:27 PM
#7
Originally posted by Dilenger4
But why would 3(5c) + c = 16c? wouldn't it be 15c2
It is, when c = 0, or c = 16/15
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Apr 18th, 2002, 04:44 PM
#8
Fanatic Member
Re: Combining like terms
Text books don't always show the right answers...
Digital-X-Treme
Contact me on MSN Messenger: [email protected]
[VBCODE]Debug.Print Round(((1097) - ((55 ^ 5 + 311 ^ 3 - 11 ^ 3) _
/ (68 ^ 5))) ^ (1 / 7), 13)[/VBCODE]
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Apr 18th, 2002, 04:46 PM
#9
Hyperactive Member
Except that in this case they have so there's not a problem there.
-Show me on the doll where the music touched you.
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Apr 19th, 2002, 12:44 AM
#10
Thread Starter
Dazed Member
Posted by Ambivalentiowa
8x3 - x2 = 8x3 + 4x2 - 5x2
So if they are equivalent then i guess i had the answer right to begin with? Thanks guys.
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Apr 19th, 2002, 01:11 AM
#11
Thread Starter
Dazed Member
Here is another one that i find hard.
6[5c + 4(d-2c)] -5c
Now i assume that i should start with the inner most grouping.
6[5c + 4d - 8c] - 5c
Now i guess i have to use the distributive property as such. But im not sure if im doing this right. 
30c + 24d - 48c - 30c
Now since the answer is supposed to be -23c + 24d i can see where 24d comes from but where does the -23c come from?
30c + -48c = -18 and -18 + (-30c) = -48c.
So i end up getting -48c + 24d
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Apr 19th, 2002, 02:06 AM
#12
Hyperactive Member
Originally posted by Dilenger4
Here is another one that i find hard.
6[5c + 4(d-2c)] -5c
Now since the answer is supposed to be -23c + 24d i can see where 24d comes from but where does the -23c come from?
30c + -48c = -18 and -18 + (-30c) = -48c.
So i end up getting -48c + 24d
6[5c+4(d-2c)]-5c
6[5c+4d-8c]-5c
6[4d-3c]-5c
24d-18c-5c
=24d-23c
You seem to have multiplied the -5c by 6 which is incorrect to do.
-Show me on the doll where the music touched you.
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Apr 19th, 2002, 05:03 AM
#13
Conquistador
JUST having a shot, seeing if i can still do it 
6[5c + 4(d-2c)] -5c
= 30c + 24(d-2c) - 5c
= 25c + 24d - 48c
= -23c + 24d
or
= 24d - 23c

lol
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Apr 19th, 2002, 07:46 AM
#14
Hyperactive Member
-Show me on the doll where the music touched you.
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Apr 19th, 2002, 02:01 PM
#15
Thread Starter
Dazed Member
Ahhhh ok. We start from the outer and work our way towards the inner.
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Apr 19th, 2002, 03:57 PM
#16
Hyperactive Member
Got any other problems? I'm better at maths than vb.
-Show me on the doll where the music touched you.
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Apr 19th, 2002, 05:44 PM
#17
Conquistador
hehehe thanks ambiva 
Dilenger, it's really just a matter of personal preference.
For this problem, i just multiplied the six out first, ambiva multiplied out the internal one first. We both got the same answer
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Apr 21st, 2002, 12:23 PM
#18
Thread Starter
Dazed Member
Posted by da_silvy
Dilenger, it's really just a matter of personal preference.
For this problem, i just multiplied the six out first, ambiva multiplied out the internal one first. We both got the same answer
Ahhhh ok, i see now. Either way it works out the same.
6[5c + 4(d -2c)] - 5c
6[5c + 4d - 8c] - 5c
30c + 24d - 48c - 5c
-23c + 24d
Thanks for the help guys.
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Apr 21st, 2002, 02:57 PM
#19
Hyperactive Member
-Show me on the doll where the music touched you.
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