SOLUTION: How do you find x,y,and z in the following equation? 5x-2y+4z=0 5x-4y-3z=-21 x-y+5z=0

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: How do you find x,y,and z in the following equation? 5x-2y+4z=0 5x-4y-3z=-21 x-y+5z=0      Log On


   

Question 499709: How do you find x,y,and z in the following equation?
5x-2y+4z=0
5x-4y-3z=-21
x-y+5z=0
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
(1)    5x - 2y + 4z =   0
(2)    5x - 4y - 3z = -21
(3)     x -  y + 5z =   0

Multiply (2) by -1 and add it to (1)

       5x - 2y + 4z =  0
      -5x + 4y + 3z = 21
     ---------------------
(4)         2y + 7z = 21

Multiply (3) by -5 and add (1)

     -5x +  5y - 25z = 0
      5x -  2y +  4z = 0
    ----------------------
(5)         3y - 21z = 0

Multiply (4) by 3 and add (5)


            6y + 21z = 63
            3y - 21z =  0
          ------------------
            9y       = 63  
                   y = 7

Substitute y = 7 in (5)

          3(7) - 21z = 0
            21 - 21z = 0           
                -21z = -21
                   z = 1

Substitute y = 7 and z = 1 in (3)

         x - (7) + 5(1) = 0
              x - 7 + 5 = 0
                  x - 2 = 0
                      x = 2

(x,y,z) = (2,7,1)

Edwin