http://tunerspec.ca/school/math1.jpg

I ended up with:

I chose option C as it is the only one that made sense to me. I can choose more than one option though... anyone have any input? Maybe I'm not solving the system correctly?

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- September 15th 2007, 12:51 PMThomasSolving Systems
http://tunerspec.ca/school/math1.jpg

I ended up with:

I chose option C as it is the only one that made sense to me. I can choose more than one option though... anyone have any input? Maybe I'm not solving the system correctly? - September 15th 2007, 01:00 PMtopsquark
- September 15th 2007, 01:11 PMThomas
Hmm... I tried just using Option A and it was wrong... I am completely stumped.

- September 15th 2007, 02:53 PMThomas
Bump!

- September 15th 2007, 05:29 PMtopsquark
- September 15th 2007, 07:28 PMThomas
- September 15th 2007, 07:52 PMticbol
Eq.(1) and Eq.(3) are the same. So there are effectively only two independent equations in the system.

Subtract (1) from (2) and we will get

4a +6b +2d = 12

Or,

2a +3b +d = 6

So,

d = 6 -2a -3b

Therefore, options A and E. ----answer. - September 15th 2007, 08:23 PMThomas
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. - September 15th 2007, 08:30 PMJhevon
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 - September 15th 2007, 08:47 PMThomas
- September 15th 2007, 08:50 PMticbol
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. - September 15th 2007, 09:01 PMThomas
Thanks for the help, ticbol. I, unfortunately, am learning matrices right now. :cool:

- September 15th 2007, 09:13 PMthe_philosipherThats odd.
Haha, it would seem we are doing the same homework, or our proffesours are getting questions from the same source. I also am having trouble with it though. Guess well just wait and see.

- September 15th 2007, 10:59 PMCaptainBlack
- September 15th 2007, 11:01 PMthe_philosipher
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).