SOLUTION: Can you help me solve this system of equations for a,b,c and d: 2a-4b+6c-2d=-6 2a-4c+d=-15 a+8b+2c+d=6 3a+6b+c+4d=2

I'll solve it part of the way step by step.  
You'll have to finish it by yourself:




b has already eliminated from the second equation,
so the best way to begin is to eliminate b from
two other equations. Multiply the 1st equation by 2
and it will have a term -8b which is opposite the
+8b in the 3rd equation, and they will cancel when
added.

4a-8b+12c-4d=-12
 a+8b+ 2c+ d=  6
----------------
5a   +14c-3d= -6

Also multiply the 1st equation by 3 and it will have 
a term -12b. Then multiply the 4th equation through
by 2 and it will have a term 12b which is opposite the
-12b, and they will cancel when added.

 6a-12b+18c-6d=-18,

 6a+12b+ 2c+8d=  4
-----------------
12a    +20c+2d=-14  

Taking these two new equations with the original equations,
we now have a system of three equations and three unknowns.



Now eliminate one of those unknowns and you'll have a
system with only two equations in two unknowns, which
you can solve.  You finish:

I'll tell you the values for b = 1/2 and d = 3.  You find
a and c by yourself.

Edwin