Use the elimination method to solve the following system of equations.

x + 3y – z = 2
x – 2y + 3z = 7
x + 2y – 5z = –21
(2, 3, 5)
(–2, 3, 5)
(2, –3, 5)
(2, 3, –5) Use the elimination method to solve the following system of equations.

x + 3y – z = 2
x – 2y + 3z = 7
x + 2y – 5z = –21
(2, 3, 5)
(–2, 3, 5)
(2, –3, 5)
(2, 3, –5) @Mathematics

Question

Use the elimination method to solve the following system of equations.

x + 3y – z = 2 
x – 2y + 3z = 7 
x + 2y – 5z = –21
(2, 3, 5)
(–2, 3, 5)
(2, –3, 5)
(2, 3, –5) Use the elimination method to solve the following system of equations.

x + 3y – z = 2 
x – 2y + 3z = 7 
x + 2y – 5z = –21
(2, 3, 5)
(–2, 3, 5)
(2, –3, 5)
(2, 3, –5) @Mathematics

jhonyy9 · Answer

assume second with 3rd and will get 2x-2z= -14 and multiply first by 2 and second by 3 and assum and will get 5x +7z = 25 so than we have 

2x-2z = -14 multiply by -1 than will be -2x +2z = 14
5x-2z = 25  

-2x+2z=14
  5x-2z=25 
------------- assume and will get 
3x = 39
 x= 39/3 = 13 
x=13 so than 5(13) -2z=25 so 65 -2z = 25 so 65-25 =2z so 40=2z so z= 40/2 = 20
z=20 than x+3y-z=2 so 13 +3y -20 =2 so 3y-7=2 so 3y=9 so y=9/3 =3

than we will get 

x=13
y=3
z=20