SOLUTION: Can I please get some assistance for this problem? Thank you. Solve the following system of three linear eqations. 2x+3y+7y=13 3x+2y-5x=-22 5x+7y-3z=-28

2x+3y+7z =  13
3x+2y-5z = -22
5x+7y-3z = -28 

1. Pick a letter to eliminate and two equations that contain that letter.

I will arbitrarily pick the letter y to eliminate and I will arbitrarily
pick the equations:

3x+2y-5z = -22
5x+7y-3z = -28

to eliminate y from.

Since the coefficients of y are 2 and 7, multiply the first
equation through by -7 and the second one through by 2 so
they will have coefficients that will cancel when equals are
added to equals:

-21x-14y+35z = 154
 10x+14y- 6z = -56
-------------------
-11x    +29z =  98

2. Next, eliminate that same letter from one of those equation
   and the equation you did not use in step 1.

I did not use 

2x+3y+7z =  13

in step 1. I will use it with

3x+2y-5z = -22

to eliminate the same letter y.

3x+2y-5z = -22
2x+3y+7z =  13

Multiply the first one through by -3 and the second
one through by 2:

-9x-6y+15z = 66
 4x+6y+14z = 26
---------------
-5x   +29z = 92

3.  Now you have a system of only two equations in two letters:

-11x+29z = 98
 -5x+29z = 92

Multiply the first equation through by -1

 11x-29z = -98
 -5x+29z =  92
--------------
  6x     =  -6
       x = -1

4. Now we switch over to substitution.  Substitute what you
got in step 3 in one of those last two equations in only two 
letters.

I will substitute -1 for x in one of those last two equations in
only two letters:

I will pick this one:
 
   -5x+29z = 92

-5(-1)+29z = 92 
     5+29z = 92
       29z = 87
         z = 3

5. Substitute those two letters in one of the original equations
that contains the first letter you eliminated.

I will substitute -1 for x and 3 for z in this original equation: 

     3x+2y-5z = -22
3(-1)+2y-5(3) = -22
     -3+2y-15 = -22
        2y-18 = -22
           2y = -4
            y = -2

The solution is (x,y,z) = (-1,-2,3).

Edwin