I have been trying to eliminate equations
Hi Smithy. We're trying to eliminate variables (possibly entire expressions, too) from equations, not eliminate equations themselves. That's what you meant to say, yes? You're trying to manipulate equations until you obtain an equation containing only one variable (to solve). Your work shows that you're trying that.
I won't have time until later this afternoon to look over all your work, and, maybe I won't have to, if you post an update. I've noticed some arithmetic errors, and some of your writing is hard to read. In fact, it may be that you can't read some of it yourself.
What is that, at the beginning of the third line? It's supposed to be 2a (from 5a-3a), but, for some reason, you wrote -2a in the fourth line. Arithmetic/sign errors are common when solving larger systems by hand, so my advice is (1) slow and steady is best and (2) separate your scratch paper from your "confirmed results" paper. That is, experiment and jot down ideas on scratch paper, but don't transfer results to your confirmed page until you've
double-checked them (i.e., did you copy what you meant to, are the signs correct, is the arithmetic verified). Cramming everything onto a sloppy scratch paper invites mistakes. Use lots of scratch paper.
In some cases, we can pick arbitrary but easy-to-work-with values for a beginning equation and use such values to evaluate a result. If we haven't goofed, then the evaluations ought to jive. If they don't, we look for the issue(s). That's way to one way to verify.
Your result of 2a+c=5 matches one of my first results. Good. We can now express c in terms of a or vice versa.
I also see b-e=1, and that matches my work, too. Good. We can now express b in terms of e or vice versa.
At this point, let's revisit the given equation involving a,b,c.
a + b + c = 6
We know that c is 5-2a. Make the substitution
a + b + 5 - 2a = 6
That yields: b - a = 1
What do we have that looks like that?
b - a = 1
b - e = 1
Aha! We see by inspection that a=e.
Then it follows that c=5-2a becomes c=5-2e.
a = e
b = e - 1
c = 5 - 2e
How about that! We know what a+b+c is, so now we can form an equation containing one variable and solve it.
Once you have the value of a variable, you may simplify each result on your confirmed page containing that variable.
Two more hints:
a + b + c = 6
a + b + c + d + e + f = 9
It's obvious what d+e+f equals.
We have two equations containing multiples of d+e+f. For example, we have
a + b + 2c + 2d + 3e + 3f = 16
That's enough stuff to make (a+b+c) and 2(d+e+f) with some leftovers, right?
(a + b + c) + c + 2(d + e + f) + e + f = 16
We have values for those two groupings. Simplify. Try the same with the other equation involving 16.
Can you work toward two equations containing only e's and f's? I think you're almost there.
?
PS: You don't need to write coefficients when they are 1. For example, write c, not 1c.
We use the factor 1 when factoring expressions: c^2+c = c(c+1)