Mathematics 32 Online
OpenStudy (anonymous):

It's pretty pathetic when I can't even do substitution for equations and everyone says I am the smart kid!! :(. Help? It's pretty pathetic when I can't even do substitution for equations and everyone says I am the smart kid!! :(. Help? @Mathematics

OpenStudy (saifoo.khan):

911?

OpenStudy (anonymous):

yes... (x+y+z=2) (x+2z=5) (2x+y-z=-1)

OpenStudy (amistre64):

thats ok, i cant do diffy qs to well and im a veritable genius lol

OpenStudy (amistre64):

we can see that x = 5-2z from the 2nd eq

OpenStudy (amistre64):

sub that into the others

OpenStudy (anonymous):

haha.

OpenStudy (amistre64):

all a system of equations is is saying that there are common values for the variables

OpenStudy (amistre64):

as such, we can substitute "quasi" values and come up with solid values

OpenStudy (amistre64):

since x has to equal 5-2z; we can simply use that in the other equations to simplify matters

OpenStudy (anonymous):

uhmm well i dont know what quasi is lol

OpenStudy (amistre64):

quasi means .... youre gonna make me look up my own words aintcha

OpenStudy (amistre64):
OpenStudy (amistre64):

having the appearance of, resembling, related to in some fashion that is not concretely established

OpenStudy (anonymous):

haha i guess so. :)

OpenStudy (amistre64):

x+y+z=2 x+2z=5 <-- x = 5-2z 2x+y-z=-1 ......................................... substitute it in (5-2z)+y+z=2 x = 5-2z 2(5-2z)+y-z=-1 ........................................ "solve" 5-2z+y+z=2 5-z+y = 2 x = 5-2z 2(5-2z)+y-z=-1 10-4z+y-z = -1 5z+y = -11 etc

OpenStudy (amistre64):

the 2 new eqs that result can be compared for either z or y now 5-z+y = 2 -z+y = -3 y = z-3 ................................. substitute this "value" since its got to hold for all ys 5z+y = -11 5z+(z-3) = -11 5z+z-3 = -11 6z = -8 z = -4/3 if im right

OpenStudy (anonymous):

okay this is what i have

OpenStudy (amistre64):

and since y = z-3, and z=-4/3 y = -4/3 -3 y = -13/3 i hate fractions they always make me worry

OpenStudy (amistre64):

and of course we had: x = 5-2z, and z=-4/3 x = 5-2(-4/3) = 5+ 8/3 = 23/3 maybe

OpenStudy (anonymous):

(x+y+z=2) (x+2z=5) (2x+y-z=-1) x=-2z+5 -2z+5+y+z=2 -z+y=-3 y=z-3 2(-2z+5)+z-3-z=-1 -4z+10+z-3-z=-1 am i doing any of this right?

OpenStudy (amistre64):

i messed it up someplace since the wolf disagrees with me ;) http://www.wolframalpha.com/input/?i=%28x%2By%2Bz%3D2%29+and++%28x%2B2z%3D5%29+and+%282x%2By-z%3D-1%29

OpenStudy (amistre64):

that looks good

OpenStudy (amistre64):

from what i can parse from it

OpenStudy (anonymous):

okay then i might have this...but i might still be coming back for more help. :)

OpenStudy (amistre64):

right here is my mess: 2(5-2z)+y-z=-1 10-4z+y-z = -1 5z+y = -11 <-- should be -5, i dropped me sign :/

OpenStudy (anonymous):

okay i figured it out lol. Thanks for the help. :).

OpenStudy (anonymous):

You will probably be getting a lot of medals from me btw lol.

OpenStudy (amistre64):

i already got lots of medals ;) but thnx anyhoos

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