Question 36286
4a - 5b -  c = -26 ------- equation 1
5a +  b + 2c = - 5 ------- equation 2
2a + 4b +  c =   6 ------- equation 3


Add eq1 and eq3 to eliminate variable c

 4a - 5b -  c = -26
+(2a + 4b +  c =   6)

 4a - 5b -  c = -26
 2a + 4b +  c =  6


We get 

6a - b  = -20    --------equation 4


Lets compare eq2 and eq3 to eliminate variable c

Multiply eq3 by 2 and subtract it from eq2

  5a +  b + 2c = - 5
-2(2a + 4b +  c =   6)

  5a +  b + 2c = - 5
 -4a - 8b - 2c = - 12

Solving we get 

   a - 7b = -17 --------equation 5


Having eliminated variable c we have got 2 equations 4 and 5

Solve 4 and 5 to get the values of a and b

6a - b  = -20    --------equation 4
 a - 7b = -17    --------equation 5


Multiply eq5 by 6 and subtract eq4 and eq5

   6a - b  = -20
-6( a - 7b = -17 )

   6a -   b  = -20
  -6a + 42b  = +102
-----------------------
        41b = 82

        b = 82/41 = 2     --- solution 1


Substitute solution 1 in either equation 4 or equation 5 to get the value of "a"
I have considered equation 4 

6a - b  = -20 
6a - 2  = -20 
6a  = -20 +2
6a  = -18
a = -3   --- solution 1


Substitute soln 1 and soln 2 in either eq 1 or eq 2 or eq 3 to get the value of "c"
I have considered equation 1

a = -3 , b = 2

4a - 5b -  c = -26
4(-3) - 5(2) -  c = -26
-12 - 10 - c = -26
-22 - c = -26
-c = -26 + 22
-c = -4

c = 4


So the answers are 
a = -3 , b = 2, c = 4


substitute in the other 2 equations to verify