anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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hartnn
  • hartnn
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 ?
anonymous
  • anonymous
Sure, but where does the z go?
anonymous
  • anonymous
only in terms of W!!!

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More answers

hartnn
  • hartnn
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
  • hartnn
do u understand how you will not be able to eliminate Z ??
hartnn
  • hartnn
i wonder what hba is typing from last 8 minutes...
hba
  • hba
@hartnn If we make more equations it is quite possible to eliminate.
hba
  • hba
I am solving it
hartnn
  • hartnn
ohh.. go ahead, surprise me ...
hartnn
  • hartnn
with 2 equations, you''l only be able to eliminate 1 variable....
hba
  • hba
Well i ended up getting their values lol and yeah @hartnn Your'e Right :)
anonymous
  • anonymous
OMG I got it. I just misread the diagram, here:|dw:1354085940249:dw|
anonymous
  • anonymous
Thats the number square associated whereas the original one was |dw:1354085985283:dw|
hba
  • hba
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
  • hba
Actually i did it.
anonymous
  • anonymous
Lol im not going to go and evaluate that xD good for you though!
hba
  • hba
Ahm,Ahm.
hartnn
  • hartnn
u subtracted incorrectly....
anonymous
  • anonymous
@hartnn :O
hartnn
  • hartnn
that was to hba....
hartnn
  • hartnn
he gets -Z=-Z after subtraction....
hba
  • hba
Where lol ?
hartnn
  • hartnn
W/3=Y-Z W+Z=X+Y - - - - ----------- W/3-W-Z=-X-Z
hba
  • hba
Oh Haha,If i wouldn't had made the mistake how could i have got the answer :P
hba
  • hba
A system with 4 variables and 2 equations.

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