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How do i solve this simple algebraic problem? |
whizkidd
Joined: May 14, 2004
Posts: > 500
From: India
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3x + 2y - z = 0 4x + 9y + 2z = 0 5x - 2z = 2 . How do i solve these three equations simultaneously to find x, y and z? I know this is very simple but i am simply not getting it. Thx! ;-)
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plasmadog
Joined: May 06, 2004
Posts: > 500
From: Beantown, India
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whizkidd
Joined: May 14, 2004
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From: India
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Er....... Thanks! :-D
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SnowBear
Joined: Oct 10, 2003
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From: Sweden
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com'n think
[addsig] |
plasmadog
Joined: May 06, 2004
Posts: > 500
From: Beantown, India
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sorry...
i posted on wap, and there might have been some problem
x=.9629
y= -0.7407
z=1.4074 |
whizkidd
Joined: May 14, 2004
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From: India
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Pleaseee do not torture me! I have spent half a day pondering over it. Please help me!
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Grimslade
Joined: Jul 25, 2004
Posts: 464
From: London
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Yes, but how's it done? I just (I confess!) spent a half-hour factoring, adding and subtracting, but I got nowhere? Anyone willing to give a quick refresher? |
methylated_spirit
Joined: Jul 07, 2004
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From: Bonnie Scotland
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do your own homework! You will never learn if someone just tells you!
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[ This Message was edited by: masseur on 2004-10-21 13:56 ] |
Johnex
Joined: Nov 26, 2002
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From: Stockholm/Sweden
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:-) do you have a ti? If so, download the polyroot finder and simultaneous equation solver. It does the trick.
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whizkidd
Joined: May 14, 2004
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From: India
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Please someone explain how it is solved. I have the answers but how to get it is what is troubling me. The answers are x=2, y=-1 and z=4.
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masseur
Joined: Jan 03, 2003
Posts: > 500
From: Sydney, London
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I don't think he wants the answers, I think he wants to know how to do it.
Its been a long time since I did simultaneous equations but... I'll try to start you off..
3x + 2y - z = 0
so by taking Z to the other side of the equal sign you get
z = 3x + 2y
now you have z expressed in terms of x and y can substitude for y in the 3rd equation
5x - 2(3x + 2y) = 2
which simplifies to
5x - 6x - 4y = 2
or simplified further
-x - 4y = 2
now swap 2 and x to give
-2 - 4y = x
so we can now substitute for x and z into the second equation to determine the value of y
once you have the numerical value of y you can supply that into the first equation (substituting z in terms of x too) and you can derive the numerical value of x leaving a simple task to find the value of Z
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Aled
Joined: Nov 25, 2002
Posts: 320
From: Wales
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Just buy a calculator that does simultaneous equations.
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masseur
Joined: Jan 03, 2003
Posts: > 500
From: Sydney, London
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can you use those in exams these days?
don't you still have to know how to do it by hand?
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whizkidd
Joined: May 14, 2004
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From: India
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CAN ANYONE HELP ME? Instead of wasting Esato's bandwidth? Please. I am getting restless. :-(
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masseur
Joined: Jan 03, 2003
Posts: > 500
From: Sydney, London
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@whizzkid, read back a few posts, I have already helped you!!
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[ This Message was edited by: masseur on 2004-10-21 14:19 ] |
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Posted: 2004-10-21 13:59
