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February 25th, 2010, 03:45 PM   #1
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Urgent help needed

Hi guys this is Dilshad again, algebra is getting tough day by day and I really don't know what's going on, we are learning to solve system of 3 linear equations. example,

8x-5y+z=15
3x+y-z=-7
x+4y+z=-3

and even these kind,

x-y =5
2y-z=1
3x+3+z=6

Is there a easy way to solve these? since I always end up with wrong answer. There must be a quick and easy way to do this, since I'm doing this for weeks for 9 hours straight and I still don't get please help

Dilshad.
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February 25th, 2010, 05:37 PM   #2
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Re: Urgent help needed

Quote:
Originally Posted by dilshad123
There must be a quick and easy way to do this, since I'm doing this for weeks for 9 hours straight and I still don't get please help

Dilshad.
Then surely with all that effort there must be something you could type in here for people to see where you might be going wrong? If there's too much typing, some people scan their written work and send it to a file hosting service for people to see it here.

Google is your friend. If you look for [solve three equations] you will come up with many sites such as this:

http://www.sosmath.com/soe/SE3001/SE3001.html
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February 26th, 2010, 03:23 AM   #3
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Quote:
Originally Posted by dilshad123
. . . 3x+3+z=6
Did you mean 3x + 3y + z = 6?
If you take anough care to avoid careless slips, you should be able to use the elimination of variables method of solution to obtain correct answers.
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February 26th, 2010, 04:13 AM   #4
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Re: Urgent help needed

Quote:
Originally Posted by dilshad123
...Is there a easy way to solve these? since I always end up with wrong answer. There must be a quick and easy way to do this, since I'm doing this for weeks for 9 hours straight and I still don't get please help...
I think substitution + elimination is already enough for me (since I haven't learned the matrix). Here's the example.
x-y =5 ------- (1)
2y-z=1 ------- (2)
3x+3+z=6 -------(3) (if the question you typed is correct)

From (1),
x = 5+y ----- (4)

Sub. (4) into (3)
3 (5+y)+3+z=6
18+3y+z = 6 --- (5)

(2)+(5)
2y-z + (18+3y+z) = 1+6
5y+18 = 7
y = -11/5

Sub. y = -11/5 into (1) , find x

Sub. y = -11/5 into (2), find z
Hope that it helps. By the way, honestly, I haven't learned to solve system of 3 linear equations.
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February 26th, 2010, 03:03 PM   #5
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Re: Urgent help needed

Thanks all for the reply, the thing is, it's when I just normally try solving I just stop when I'm moving the variables example making positive 3's into negative 3's moving to the other side, then I stop totally since I don't see how I can go any further. The mind just goes blank at me.
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February 26th, 2010, 03:37 PM   #6
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Re: Urgent help needed

Quote:
Originally Posted by dilshad123
Thanks all for the reply, the thing is, it's when I just normally try solving I just stop when I'm moving the variables example making positive 3's into negative 3's moving to the other side, then I stop totally since I don't see how I can go any further. The mind just goes blank at me.
Then I hate to tell you, but you really need to go back to earlier, simpler examples with two variables, or even rearranging one equation or formula until you master that concept. Later concepts rely so much upon earlier skills, and that's the truth of it. It is difficult to teach an entire course of study here. Also, as is stated often, it is even more difficult to see where you make errors if those errors are not to be seen. Why can you not do some typing when you expect that of those who are willing to help you in return? If someone solves it for you, doing all the work, it is then just one more example, and you say you do not understand what you have already seen that will be exactly the same process.
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