SOLUTION: 3. Solve for X, Y, and Z in the following systems of three equations using either substition or elimination methods: a. X + 2Y + Z = 22 X + Y = 15 3X + Y + Z = 37 b. 1

I'll do the second one only:

10X +   Y +  Z = 603
 8X +  2Y +  Z = 603
20X - 10Y - 2Z =  -6

Pick any two equations and a letter to eliminate from them.
I could pick any two equations and any letter, but you might as
well pick the easiest letter to eliminate and the two easiest
equations to eliminate it from.  So I will pick the first two
equations and eliminate Z from them by multiplying the first one
by -1 and adding it to the second equations:

10X +  Y + Z =  603
-8X - 2Y - Z = -603
-------------------
 2X -  Y     =   0

Now use one of those equations with the third equation and eliminate
the same letter, Z.  I'll multiply the first original equation by 2
and add it to the third equation to make the Z's cancel

16X +  4Y + 2Z = 1206
20X - 10Y - 2Z =   -6
---------------------
36X -  6Y      = 1200

and we can divide that through by 6 to make it easier:

6X - Y = 200

Next we put those two resulting equations together:

2X - Y =   0
6X - Y = 200

Multiply the first one of those by -1 so the Y's will cancel:

-2X + Y =   0
 6X - Y = 200
-------------
 4X     = 200
      X = 50

Substitute X = 50 into either one of those equations. I'll
pick the first one:

   -2X + Y = 0
-2(50) + Y = 0
  -100 + Y = 0
         Y = 100

Now you have two of the unknowns, so you pick any one
of the original three equations and substitute those two
values.  I'll pick the second original equation:

      8X +  2Y  + Z = 603
 8(50) + 2(100) + Z = 603
      400 + 200 + Z = 603
            600 + Z = 603
                  Z = 3

(X,Y,Z,) = (50,100,3)

Edwin