Question

.-.

Answer 1

and if y'all wait a few minutes, i can post the work i've done so far

Answer 2

camera's charging :l

Answer 3

i'd really like to see where i messed up. i got x = -3/5, but it's not right

Answer 4

\[(i)\qquad2x+2y-z=1\]\[(ii)\qquad2x+3y+z=8\]\[(iii)\qquad3x-y-2z=-4\]

\[-(i)\qquad-2x-2y+z=-1\]

\[(ii)+(iii)-(i)\]\[\quad=(2x+3y+z)+(3x-y-2z)+(-2x-2y+z)=8+(-4)+(-1)\]
solve for x

Answer 5

you should find x=1

Answer 6

ok, one second...i'll post the picture...i dunno how i got -3/5...

Answer 7

this is what i did...can you tell me what i did wrong?

Answer 8

how would i get x = 1? i don't understand what you did :<

Answer 9

you left off a negative sign

Answer 10

omg. realleh.

Answer 11

ok...lemme see what i get when i fix it

Answer 12

\[\ddot\smile\]

Answer 13

x = 1! omg!

Answer 14

thanks so much!

Answer 15

good, keep going

Answer 16

oh right, because i need y and z. sigh.

Answer 17

y = 1 too?

Answer 18

thats right,

Answer 19

z = 3?

Answer 20

\[\Huge \color{red}{\checkmark\checkmark\checkmark}\]

Answer 21

woohoo! thank you!!