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

Steps:

isolate " x "  for -2x + 2y  + 3z  = 0 ;    x = -  ( - 3z - 2y ) / 2

Subsititute : x = -  ( - 3z - 2y ) / 2

- 2 * (  - ( - 3z - 2y ) / 2 ) - y + z = - 3

2 * ( -  ( - 3z - 2y ) / 2 ) + 3y  + 3z  =5

isolate " z " for  :  - 2  ( -  ( - 3z - 2y ) / 2 ) - y+ z = - 3 ;  z = - ( - 3 +  3y / 2 )

subsititute  z = - ( - 3 + 3y / 2 ) :

2* ( - ( -3* ( - ( - 3 +3y / 2 ) - 2y / 2  ) + 3y  + 3  ( - ( - 3 + 3y / 2 ) = 5

isolate y for 2 ( -( - 3 ) - 3 + 3y / 2 ) - 2y / 2 ) + 3y + 3 ( - ( - 3 + 3y / 2 ) = 5

y = 1

for z =  - ( - 3+ 3y ) / 2

subsititute y = 1

z = - ( - 3 + 3  * 1 ) / 2

z =  0

For x  = - ( - 3z - 2y ) / 2

subsititute y = 1

subsititute z = 0

x = -  ( - 3 *0 - 2 * 1 ) / 2

x =  1

x = 1  , y =  1  , z = 0

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]

c. (3,2,6) is the answer.