Use gauss-jordan elimination to solve the following linear system.
x – 2z = 9
6x – 2y – 5z = 29
–5x + 5y + 3z = –14

a. (–5,–3,0)
b. (3,2,–3)
c. (5,3,–5)
d. (3,6,–4)

x – 2z = 9 so x = 2z + 9
6x – 2y – 5z = 29
–5x + 5y + 3z = –14

substitute x = 2z + 9 into 6x – 2y – 5z = 29 and –5x + 5y + 3z = –14

6(2z + 9) – 2y – 5z = 29
12z + 54 - 2y -5z =29
-2y + 7z = - 25 (1st equation)

–5(2z + 9) + 5y + 3z = –14
-10z - 45 + 5y + 3z = -14
5y - 7z = 31 (2nd equation)

multiply (1st) equation by 5 and (2nd) equation by 2
-10y + 35z = - 125
 10y - 14z = 62
add
21z = - 63
 z = -3

substitute z = - 3 into x = 2z + 9

x = 2z + 9
x = 2(-3) + 9 
x = 3

substitute x =  3 and z = 3 into 6x – 2y – 5z = 29

6x – 2y – 5z = 29
6(3) – 2y – 5(-3) = 29
18 - 2y + 15 = 29
-2y + 33 = 29
-2y = -4
  y  = 2
x = 3, y = 2 and z = -3

answer
B. (3, 2, –3)

A’b’c’ are the coordinates to the new location of the transformed triangle abc. find the distance between the coordinates points to use for your calculations.