Question 116081
<pre><font size = 6><b>
 2a + 3b - 4c + 6d =  6 
 7a - 5b - c       = -7 
13a - 9b           =  6 
           dē - 2d = -1


Solve the last equation first, since
that is the only equation that has 
just one unknown:

       dē - 2d = -1
   dē - 2d + 1 =  0
(d - 1)(d - 1) =  0
d - 1 = 0 or d =  1

Substitute d = 1 into the first 
equation:

  2a + 3b - 4c + 6d = 6
2a + 3b - 4c + 6(1) = 6
   2a + 3b - 4c + 6 = 6
       2a + 3b - 4c = 0

So now the system of equations is:

 2a + 3b - 4c =  0
 7a - 5b -  c = -7 
13a - 9b      =  6

Can you solve that system? If not,
post again asking how.

Answer:  a = 6, b = 8, c = 9 

and earlier we found d = 1.

Edwin</pre>