SOLUTION: How do you solve this problem? x+y+z-w=2 3x+y-z+w=8 x-5y+2z+w=1 -x-y+z+3w=4

Question 418129: How do you solve this problem?
x+y+z-w=2
3x+y-z+w=8
x-5y+2z+w=1
-x-y+z+3w=4
Answer by Edwin McCravy(20056) (Show Source): You can put this solution on YOUR website!


There are many ways to solve it.  You didn't specify which way.
I arbitrarily picked elimination.  Which method are you studying?
Gaussian elimination?  Cramer's rule? If you're studying another 
method, then post again telling what method.  We aren't mind
readers.

 x +  y +  z -  w = 2
3x +  y -  z +  w = 8
 x - 5y + 2z +  w = 1
-x -  y +  z + 3w = 4

1. Pick a letter to eliminate.  I'll pick w
2. Pick two equations to eliminate it from.  I pick the eq. 1 and eq. 2 

 x +  y +  z -  w =  2
3x +  y -  z +  w =  8
----------------------
4x + 2y           = 10

It can be divided through by 2

           2x + y = 5

3. Pick two more equations to eliminate z from.  
   I pick the eq. 2 and eq. 4

3x +  y -  z +  w =  8
-x -  y +  z + 3w =  4
----------------------
2x           + 4w = 12

It can also be divided through by 2

           x + 2w = 6

3. Pick two more equations to eliminate z from. 
   But we must be sure this time to pick the equation we 
   haven't picked before. 
   I pick the eq. 2 and eq. 3
  
3x +  y -  z +  w = 8
 x - 5y + 2z +  w = 1

We must multiply the top one by 2 to make the z's cancel:

6x + 2y - 2z + 2w = 16
 x - 5y + 2z +  w =  1
----------------------
7x - 3y      + 3w = 17

Now we have this system

2x +  y      =  5
 x      + 2w =  6
7x - 3y + 3w = 17

Since y is already eliminated from the 2nd eq., we will
eliminate y from the 1st and 3rd.

2x +  y      =  5
7x - 3y + 3w = 17
 
We must multiply the top one by 3 to make the y's cancel:

 6x + 3y      = 15
 7x - 3y + 3w = 17
------------------
13x      + 3w = 32

Now we put that with the 2nd eq. and we have this system:

       x + 2w =  6
     13x + 3w = 32

We must multiply the top one by -13 to make the x's cancel:

    -13x - 26w = -78
     13x +  3w =  32
   -----------------
          -23w = -46
             w =  2

Substitute w = 2 in

       x + 2w = 6
     x + 2(2) = 6
        x + 4 = 6
            x = 2

Substitute x = 2 in

       2x + y = 5
     2(2) + y = 5 
        4 + y = 5
            y = 1

Substitute x = 2, y = 1, and w = 2 in

x + y + z - w = 2
2 + 1 + z - 2 = 2
        1 + z = 2
            z = 1

Solution: (x,y,z,w) = (2,1,1,2)

Edwin

RELATED QUESTIONS
How would you solve the 4x4 w+x+y+z=2, 2w-x-y+2z=7, 2w+3x+2y-z=-2,... (answered by Fombitz)
solve Linear sestems when w=1 w+2x+2y+z=-2 w+3x-2y-z=-6 -2w-x+3y+3z=6 w+4x+y+2z=-14 (answered by Fombitz)
Please help me solve this 4x4 equation w+x+y+z=2 2w-x-y+2z=7 2w+3x+2y-z=-2... (answered by stanbon,MathLover1)
what are the values of w,x,y, and z in these equations? 1. w + x + y + z = 2 2. 2w -... (answered by ankor@dixie-net.com)
Express the solution in the form(x,y,z,w) 3x+y-z+2w=9 x+y+2z-w=10 x-y-z+3w=-2... (answered by asha)
Solve the system using matrices. Express the solution set as {w,x,y,z} w - x + y + z = (answered by Fombitz)
I need step by step instructions on how to solve the following: x + 2y - 2z + 4w = 1 (answered by jim_thompson5910)
x+y+3z+2w=13 x-2y+2z+w=8... (answered by Alan3354)
x+y+z=-2 -3x-5y-z=6 2y-3z=2 How do you solve... (answered by Alan3354)