Question 370091
Use the elimination method to solve the following system of equations. 
x + 3y – z = 2
x – 2y + 3z = 7
x + 2y – 5z = –21 


x + 3y -  z =   2
x - 2y + 3z =   7
x + 2y - 5z = -21
pick a variable to eliminate, choosing x since all the x coefficients are the same, multiply the 1st equation by -1 and add it to the 2nd
-x - 3y +  z =  -2
0x - 5y + 4z =   5
 x + 2y - 5z = -21
add 1st equation to 3rd equation
-x - 3y +  z =  -2
0x - 5y + 4z =   5
0x -  y - 4z = -23
now all the coefficients on the x's are zeroed out save one,
multiply the 3rd equation by -5 and add the 2nd equation to it
-x - 3y +   z =  -2 --> -x - 3y +   z =  -2
0x - 5y +  4z =   5 --> 0x - 5y +  4z =   5
0x + 5y + 20z = 115 --> 0x + 0y + 24z = 120
multiply the 1st equation by 5, and the 2nd equation by -3, and add the 2nd equation to the first
-5x - 15y +  5z =  -10 --> -5x +  0y  -  7z =  -25
 0x + 15y - 12z =  -15 -->  0x + 15y  - 12z =  -15
 0x + 0y  + 24z =  120 -->  0x +  0y  + 24z =  120
multiply the 2nd equation by 2, add the 3rd equation to the 2nd
-5x +  0y  -  7z =  -25 --> -5x +  0y  -  7z =  -25
 0x + 30y  - 24z =  -30 -->  0x + 30y  +  0z =   90
 0x +  0y  + 24z =  120 -->  0x +  0y  + 24z =  120
now we have 2 of the x coefficients zeroed out and 2 of the y coefficients zeroed out, now for the z's, multiply the first equation by 8, and the 3rd equation by 7/3, add the 3rd equation to the first
-40x +  0y  - 56z = -200 --> -40x +  0y  +  0z =  80
  0x + 30y  +  0z =   90 -->   0x + 30y  +  0z =  90 
  0x +  0y  + 56z =  280 -->   0x +  0y  + 56z = 280
now divide the 1st equation by -40, the 2nd equation by 30,
and the 3rd equation by 56
 x + 0y + 0z = -2
0x +  y + 0z =  3
0x + 0y +  z =  5
so x = -2, y = 3, z = 5
check:
x + 3y -  z =   2 --> -2 + 9 -  5 =  7 -  5 =   2, yes
x - 2y + 3z =   7 --> -2 - 6 + 15 = -8 + 15 =   7, yes
x + 2y - 5z = -21 --> -2 + 6 - 25 =  4 - 25 = -21, yes