Ahh, it never clued in for me that EQN #1 and EQN #3 were the same, stupid!
The answer ends up being A,B, and E.
I'm not sure if your way works as well, but once you made me realize they were the same, I removed EQN #3 and did what I had been doing before.
Here's what I did...
We have:
Therefore:
Then I can solve the first equation for c and substitute it into the second equation, which satisfies Option E. Option A is satisfied by looking at the matrix. Option B is also satisfied.
do not remove the 3rd equation, you need it to know how to set up your parameters. if two equations are the same, it means you can use one to cancel everything in the other, so you end up with a row of zeros, so you should start with:
now find the reduced row echelon form of that
I never did like matrices. I avoid doing them.
My "way" is the garden-variety methods of solving simultaneous equations---by substitution, elimination, etc.
Yes, option B is correct too.
Plug a = -1 into (1),
-7 +3b +c +d = 5
3b +c +d = 12 --------(5)
Plug a = -1 into
d = 6 -2a -3b,
d = 6 +2 -3b
d = 8 -3b
Substitute that into (5),
3b +c +8-3b = 12
c = 4
So, option B.
I would have guessed that but the question is slightly different in how the possible answers are listed, theyre the same only in a different order. Also the image he used is in a different format, which suggests hes using a different program and the system I am using is the only way we can get our question, so its from two different sources and different professours (unless our professour is doing both seperately, but that seems unlikely, just a common question i think).