if i am trying to find y and it ends up being 0=0 what do i do? if i am trying to find y and it ends up being 0=0 what do i do? @Mathematics

cry

0=0 is always true so its all solutions

or rather infinite solutions

lol i already did last night.

okay so if i am looking for x y and z and y is 0=0 then its x=5 z=-2 and y=infinite solutions?

or the whole problem is infinite solutions?

dunno, depends on how right you did the math from the original problem

im pretty sure its all right.

whats the original so that I can compare to it?

solve by substitution (x-4y+z=6) (2x+5y-z=7) (2x-y-z=1)

x- 4y+z = 6 2x+5y -z = 7 2x - y -z = 1 2x= 7-5y + z (7-5y + z) - y -z = 1 7-6y = 1 -6y = -6 y = 1 you sure you did it right?

haha okay i did do it wrong. Thanks lol.

x- 4y+z = 6 2x+5y -z = 7 2x - y -z = 1 x- 4+z = 6 2x+5 -z = 7 2x - 1 -z = 1 lol ... good luck with it ;)

Thanks. :).

okay so i am confused again!!! :(.

lol ... what is it know?

okay so i got y=1 then i got confused...:(. i feel soooooo freaking stupid!

ok, we know y = 1, or at least it has to for this to work out; so lets go ahead and use it: x- 4+z = 6 2x+5 -z = 7 2x - 1 -z = 1 x+z = 10 2x -z = 2 2x -z = 2 the last 2 equations are the same so we can just use one of them right? x+z = 10 ; x = 10-z 2x -z = 2 2(10-z)-z=2 20 -2z -z=2 -3z = -18 z = 6

using the first eq; x+z = 10; and z=6, x has to be 4

dbl chk

x- 4y+z = 6 4 1 6 ------------ 4 - 4 + 6 = 6 , good 2x+5y -z = 7 4 1 6 ------------- 8 + 5 - 6 =7; good 2x - y -z = 1 4 1 6 ------------ 8 - 1 - 6 = 1 ; good

okay now i feel really stupid because that ended up being simple...lol thanks. Again. :).

;)

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