Solve the system of elimation
-2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z=5
Need All Steps Written Out

Question

Solve the system of elimation 
-2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z=5 
Need All Steps Written Out

unklerhaukus · Answer

if you add the third equation to either of the first two , the x's will cancel out

anonymous · Answer

Okay what do i do after that

unklerhaukus · Answer

What did you get?

anonymous · Answer

2y+3z=0 and -y+z=-3 Right?

unklerhaukus · Answer

Add the third equation to the first equation like this 

[2x+3y+3z   =   5 ]
                     + 
(-2x+2y+3z  =  0  )

[2x+3y+3z]  + (-2x+2y+3z) =   [5] + (0)

anonymous · Answer

nvm it would be 3y+3z=5 and -y=z+-3 Right

anonymous · Answer

oh okay then what do i do

unklerhaukus · Answer

[2x+3y+3z]  + (-2x+2y+3z) =   [5] + (0) 
2x+3y+3z        -2x+2y+3z    =   5 + 0

simplify this, 

hopefully all the x's will go

anonymous · Answer

so after you take out the xs your left with 3y+3z+2y+5z=5+0 Right

unklerhaukus · Answer

simplify

anonymous · Answer

then your left with 1y+2z=5 Right?

unklerhaukus · Answer

try that again 

2x+3y+3z        -2x+2y+3z    =   5 + 0

(2-2)x  +  (3+2)y  +  (3+3)z   =   5 + 0

anonymous · Answer

0x+1y+6z=5

unklerhaukus · Answer

almost
(2-2)x  +  (3+2)y  +  (3+3)z   =   5 + 0
0x        +  (3+2)y  +        6z   =   5