Find the values of X.Y.Z
2x+y+z=5
x-y+2z=4
x+y+z=4
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OpenStudy (anonymous):
@lgbasallote
OpenStudy (anonymous):
@Hero help
OpenStudy (lgbasallote):
add equation 2 and equation 3
OpenStudy (anonymous):
how?
OpenStudy (lgbasallote):
x - y + 2z = 4
x + y + z = 4
what's x + x?
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OpenStudy (anonymous):
x^2?
OpenStudy (lgbasallote):
nope that is x times x
OpenStudy (anonymous):
2x?
OpenStudy (lgbasallote):
better
OpenStudy (lgbasallote):
now what's -y + y
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OpenStudy (anonymous):
0
OpenStudy (lgbasallote):
good..what's 2z + z?
OpenStudy (anonymous):
3z
OpenStudy (lgbasallote):
and 4+4?
OpenStudy (anonymous):
8
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OpenStudy (lgbasallote):
therefore
x -y + 2z = 4
x + y + z = 4
------------
2x + 3z = 8 <---let this be equation 4 (we set it aside for now)
OpenStudy (lgbasallote):
now add equation 1 and 2
OpenStudy (anonymous):
okay i did that and i got 3x+3x=9 do i combine that with equation 4?
@igbasllote
OpenStudy (lgbasallote):
dont you mean 3x + 3z = 9?
OpenStudy (lgbasallote):
coz im reading 3x + 3x = 9..my eyes could be blurred though
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OpenStudy (anonymous):
yes
OpenStudy (lgbasallote):
well then very good you combine it with equation 4
3x + 3z = 9
2x + 3x = 8
the question is...do you add or subtract?
OpenStudy (anonymous):
add?? :/
OpenStudy (lgbasallote):
let's try...
3x + 3z = 9
2x + 3x = 8
-----------
5x + 6z = 17 <---you still have two variables
remember the point of this thing is to lessen the number of variables...
OpenStudy (lgbasallote):
so since adding didnt work you subtract
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OpenStudy (anonymous):
ahh okay
OpenStudy (lgbasallote):
so subtract
3x + 3x = 9
2x + 3x = 8
what do you get as difference?
OpenStudy (anonymous):
1x or X and 0z and 1
OpenStudy (anonymous):
(@lgbasallote lol you have been typing 2x + 3x =8, think it's supposed to 2x + 3z = 8)
OpenStudy (anonymous):
@nphuongsun93 right! lol
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OpenStudy (lgbasallote):
lol im making the same mistake =)))))
OpenStudy (lgbasallote):
anyway @sharee2012 you're right x = 1
OpenStudy (anonymous):
ahh sleep calls lol
OpenStudy (anonymous):
thank you :D very much for the help
OpenStudy (lgbasallote):
now substitute x = 1 into equation 4 or 5 whichever you prefer to get z...then substitute x and z into equation 1,2 or 3
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OpenStudy (lgbasallote):
thank you for being a participative learner <tips hat>
OpenStudy (anonymous):
im a little confused
OpenStudy (lgbasallote):
confused where?
OpenStudy (anonymous):
the pluggin in to solve the last part -_-
OpenStudy (lgbasallote):
have you found out what z is yet?
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OpenStudy (anonymous):
is z 0?
OpenStudy (anonymous):
@igbasallote?
OpenStudy (anonymous):
Not an i, it's an L. :)
Z is not zero.
3x+3z = 9
You've already found x = 1, so substitute the value of x:
3(1)+3z=9
OpenStudy (lgbasallote):
lol sorry i was not able to come...you tagged the wrong account it's LGBASALLOTE. IGBASALLOTE is a different user lol
OpenStudy (lgbasallote):
anyway thanks for coming on my behalf @Wired
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OpenStudy (anonymous):
oh okay sorry im so tired lol and okay i get it @Wired yes thank u very much :D