Areesha.1D
1) 2W=3X and W+Z=X+Y what is Y in terms of W?
(i caught myself stuck on this when reviewing, when i had gotten it right the first time around :S)
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hartnn
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you'll be able to find Y in terms of W and Z
to get 3X in 2nd equation, you multiply W+Z=X+Y by 3
so that you can replace '3X' by '2W' (from first equation.)
what u get ?
Areesha.1D
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Sure, but where does the z go?
Areesha.1D
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only in terms of W!!!
hartnn
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thats what i am saying, you won't be able to find it ONLY in terms of W , it seems impossible to me..... unless more equations are given.....
hartnn
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do u understand how you will not be able to eliminate Z ??
hartnn
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i wonder what hba is typing from last 8 minutes...
hba
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@hartnn If we make more equations it is quite possible to eliminate.
hba
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I am solving it
hartnn
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ohh.. go ahead, surprise me ...
hartnn
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with 2 equations, you''l only be able to eliminate 1 variable....
hba
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Well i ended up getting their values lol and yeah @hartnn Your'e Right :)
Areesha.1D
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OMG I got it. I just misread the diagram, here:|dw:1354085940249:dw|
Areesha.1D
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Thats the number square associated whereas the original one was |dw:1354085985283:dw|
hba
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X=2W/3
W+Z=2W/3+Y
W-2W/3=Y-Z
W/3=Y-Z
W+Z=X+Y
- - - -
-----------
W/3-W-Z=X-Z
W/3-W=2W/3-Z
W/3-W-2W/3=Z
W-3W-2W/3=Z
-4W/3=Z
W+Z=X+Y
W-4W/3=2W/3+Y
hba
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Actually i did it.
Areesha.1D
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Lol im not going to go and evaluate that xD good for you though!
hba
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Ahm,Ahm.
hartnn
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u subtracted incorrectly....
Areesha.1D
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@hartnn :O
hartnn
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that was to hba....
hartnn
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he gets -Z=-Z after subtraction....
hba
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Where lol ?
hartnn
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W/3=Y-Z
W+Z=X+Y
- - - -
-----------
W/3-W-Z=-X-Z
hba
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Oh Haha,If i wouldn't had made the mistake how could i have got the answer :P
hba
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A system with 4 variables and 2 equations.