SOLUTION: Instructions were to "Solve the system of linear equations" (There is a bracket( { ) on one side of all three equations) -x + y + 2z = 1 2x + 3y + z = -2 5x + 4y + 2z = 4

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

If we have this system of equations, 

Ax + By + Cz = D
Ex + Fy + Gz = H
Ix + Jy + Kz = L

We want to change it to a system that has this form:

Mx + Ny + Pz = Q
     Ry + Sz = T
          Uz = V 

So we want to get 

eq (1)  -x +  y + 2z =  1
eq (2)  2x + 3y +  z = -2
eq (3)  5x + 4y + 2z =  4

that way.  To eliminate the 2x, we multiply eq (1) thru by 2
and add it to 1 times eq (2).  

2[-x +  y + 2z =  1]   --- >   -2x + 2y + 4z =  2
1[2x + 3y +  z = -2]   --- >    2x + 3y +  z = -2
  5x + 4y + 2z =  4            ———————————————————
                                     5y + 5z =  0

We observe that we can have a simpler equation by dividing
this equation through by 5:           y +  z =  0 

So we write that equation as the new eq (2). Now we have

eq (1)  -x +  y + 2z =  1
eq (2)        y +  z =  0
eq (3)  5x + 4y + 2z =  4

Next we eliminate the 5x. To do that, we multiply eq (1) 
thru by -5 and add it to 1 times eq (3).  

5[-x +  y + 2z = 1]    --- >   -5x + 5y + 10z = 5
        y +  z = 0   
1[5x + 4y + 2z = 4]    --- >    5x + 4y +  2z = 4         
                                     9y + 12z = 9

We observe that we can have a simpler equation by dividing
this equation through by 3:          3y +  4z = 3 

So we write that equation as the new eq (3). Now we have

eq (1)  -x +  y + 2z = 1
eq (2)        y +  z = 0
eq (3)       3y + 4z = 3

Finally we eliminate the 3y. To do that, we multiply eq (2) 
thru by -3 and add it to 1 times eq (3).  

   -x +  y + 2z = 1    
-3[      y +  z = 0]    --- >        -3y - 3z = 0      
 1[     3y + 4z = 3]    --- >         3y + 4z = 3         
                                            z = 3

So we write that equation as the new eq (3). Now we have

eq (1)  -x +  y + 2z = 1
eq (2)        y +  z = 0
eq (3)             z = 3

Now all we need to do is do "back substitution".

Since eq (3) tells us that z = 3, we replace z by 3 in eq (2)

              y + z =  0
              y + 3 =  0
                  y = -3

Now that we know both y and z, we can substitute those in eq (1)
and find x

              -x + y + 2z =  1
         -x + (-3) + 2(3) =  1
               -x - 3 + 6 =  1 
                   -x + 3 =  1
                       -x = -2
                        x =  2

So the solution is (x, y, z) = (2, -3, 3)

Edwin
AnlytcPhil@aol.com