SOLUTION: can you help me solve this system of linear equations that has four variables by showing me the steps on how to do it with work too. w+x-y+z=0 w-2x-2y-z=-5 w-3x-y+z=4 2w-x-y+

Question 767198: can you help me solve this system of linear equations that has four variables by showing me the steps on how to do it with work too.
w+x-y+z=0
w-2x-2y-z=-5
w-3x-y+z=4
2w-x-y+3z=7

Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
Better to use matrices and their row operations.
1 1 -1 1 0
1 -2 -2 -1 -5
1 -3 -1 1 4
2 -1 -1 3 7

Use any multiple of row 1 to eliminate the w in all rows below.
1 1 -1 1 0
0 -3 -1 -2 -5
0 -4 0 0 4
0 -3 1 1 7

From there, only giving descriptions of the process:
switch row 2 with row 3;
3*row2, 4*row3, 4*row4;
subtract multiples of row2 to below rows;
simplify row2,3,4 dividing to get their lowest terms;

1 1 -1 1 0
0 1 0 0 -1
0 0 1 2 8
0 0 2 2 13

Almost there...
Next, subtract 2*row3 from row4:
Let row3=row3+row4.

The matrix is now:
---------------------
1 1 -1 1 0
0 1 0 0 -1 0
0 0 1 0 5
0 0 0 -2 -3
---------------------

That shows nearly the whole solution. Remember, the rows are columns are in order, w x y z Konstant. You CAN READ EQUATIONS from the matrix and get the variables values. Experts know how to do even further operations on the matrix, but you have enough without doing that.

The row3 tells you
The row4 tells you -2z=-3, so get 2z=3 meaning .
Row2 tells you x-z=0, meaning x=z, meaning because you already found z.
Row1 gives you w+x-y+z=0, and you can substitute what you already found:
w+3/2-5+3/2=0
w-5=0

-------------------------------------------
SUMMARY OF SOLUTION RESULT:_____|
w=5; x=3/2; y=5; z=3/2________________|
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