Question

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

Answer 1

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

Answer 2

Okay what do i do after that

Answer 3

What did you get?

Answer 4

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

Answer 5

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)

Answer 6

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

Answer 7

oh okay then what do i do

Answer 8

[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

Answer 9

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

Answer 10

simplify

Answer 11

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

Answer 12

try that again

2x+3y+3z -2x+2y+3z = 5 + 0

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

Answer 13

0x+1y+6z=5

Answer 14

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