x +  y +  z = 3
 2x -  y +  z = 2
 4x - 2y + 3z = 5

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



The idea is to use row operations to get it to this form 

Add -2×row1 to 1×row2 to get a 0 where the 2 is on the left
of row2:

 = 

Add -4×row1 to 1×row3 to get a 0 where the 4 is on the left
of row3:

 = 

Add 1×row2 to 3×row1 to get a 0 where the 1 is on row1 2nd elementt

 = 

Add -2×row2 to 1*row3 to get a 0 where the -6 is:

 = 

Add -2×row3 to 1*row1 to get a 0 where the 2 is:

 = 

Add 1×row3 to 1*row2 to get a 0 where the -1 is:

 = 

We have all the 0's placed, so all we need do is get the 1's

Get a 1 where the first 3 on row1 is by dividing R1 by 3
Get a 1 where the first -3 on row2 is by dividing R2 by -3
 = 

This is the row-reduced echelon form.  To get the solution, translate
it into a system of 3 equations and 3 variables:

 1x + 0y + 0z = 1
 0x + 1y + 0z = 1
 0x + 0y + 1z = 1

or

x = 1
y = 1
z = 1

So the solution is (x,y,z) = (1,1,1)

Edwin