SOLUTION: Can you help solve these functions with 3 variables? 4a + 2b - c = 5 2a+ b - 5c = -11 a - 2b + 3c = 6

eq. 1       4a + 2b -  c =   5
eq. 2       2a +  b - 5c = -11
eq. 3        a - 2b + 3c =   6

You can eliminate b immediately from eqs. 1 and 3 by adding them
vertically term-by-term:

eq. 1       4a + 2b -  c =   5
eq. 3        a - 2b + 3c =   6
           -------------------
eq. 4       5a      + 2c =  11

So we must eliminate be from another pair of equations:

We multiply eq. 2 through by 2 

            4a + 2b - 10c = -22

so we can add eq. 3 to it and eliminate b from that pair:

            4a + 2b - 10c = -22
eq. 3        a - 2b +  3c =   6
            -------------------
eq. 5       5a      -  7c = -16

Now we take eqs. 4 and 5 together as a system of two equations in two
unknowns:

eq. 4       5a + 2c =  11
eq. 5       5a - 7c = -16
   
We multiply eq. 5 by -1 

           -5a + 7c = 16

so we can add eq. 4 to it to eliminate "a":

           -5a + 7c = 16
eq. 4       5a + 2c = 11
          ----------------
                 9c = 27
                  c = 3
   
Substitute c = 3 in eq. 4

          5a + 2(3) =  11
             5a + 6 = 11
                 5a = 5
                  a = 1

Substitute c = 3 and a = 1 in eq. 3

    (1) - 2b + 3(3) =   6
         1 - 2b + 9 = 6
            10 - 2b = 6
                -2b = -4
                  b = 2

So the solution is (a,b,c) = (1,2,3)

Edwin