# Solving Systems

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Sep 15th 2007, 12:51 PM
Thomas
Solving Systems
http://tunerspec.ca/school/math1.jpg

I ended up with:

$\displaystyle a = \frac{-1}{5}$
$\displaystyle b = \frac{32}{15} - \frac{1}{3}d$
$\displaystyle c = 0$

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?
• Sep 15th 2007, 01:00 PM
topsquark
Quote:

Originally Posted by Thomas
http://tunerspec.ca/school/math1.jpg

I ended up with:

$\displaystyle a = \frac{-1}{5}$
$\displaystyle b = \frac{32}{15} - \frac{1}{3}d$
$\displaystyle c = 0$

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?

Edit (Take 2):

Note that equations 1 and 3 are the same (equation 3 is equation 1 mulitplied by 3.) So we really have two equations in five unknowns, so it should be answer D.

-Dan
• Sep 15th 2007, 01:11 PM
Thomas
Hmm... I tried just using Option A and it was wrong... I am completely stumped.
• Sep 15th 2007, 02:53 PM
Thomas
Bump!
• Sep 15th 2007, 05:29 PM
topsquark
Quote:

Originally Posted by Thomas
Hmm... I tried just using Option A and it was wrong... I am completely stumped.

It isn't D?

-Dan
• Sep 15th 2007, 07:28 PM
Thomas
Quote:

Originally Posted by topsquark
It isn't D?

-Dan

Nope... well it might be D, but it will include one of the other answers as well.

These couple questions I'm having trouble with are beginning to be ridiculous.
• Sep 15th 2007, 07:52 PM
ticbol
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.
• Sep 15th 2007, 08:23 PM
Thomas
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:$\displaystyle \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;7 & 3 & 1 & 1 & 5 \;\\\;11 & 9 & 1 & 3 & 17\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}$
$\displaystyle \frac{R1}{7} \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & \frac{3}{7} & \frac{1}{7} & \frac{1}{7} & \frac{5}{7} \;\\\;11 & 9 & 1 & 3 & 17\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}$
$\displaystyle R2 - 11\times R1 \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & \frac{3}{7} & \frac{1}{7} & \frac{1}{7} & \frac{5}{7} \;\\\;0 & \frac{30}{7} & \frac{-4}{7} & \frac{10}{7} & \frac{64}{7}\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}$
$\displaystyle R2\times\frac{7}{30} \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & \frac{3}{7} & \frac{1}{7} & \frac{1}{7} & \frac{5}{7} \;\\\;0 & 1 & \frac{-2}{15} & \frac{1}{3} & \frac{32}{15}\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}$
$\displaystyle R1 - \frac{3}{7}R2 \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & \frac{1}{5} & 0 & \frac{-1}{5} \;\\\;0 & 1 & \frac{-2}{15} & \frac{1}{3} & \frac{32}{15}\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}$

Therefore:
$\displaystyle a + \frac{1}{5}c = \frac{-1}{5}$
$\displaystyle b - \frac{2}{15}c + \frac{1}{3}d = \frac{32}{15}$

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.
• Sep 15th 2007, 08:30 PM
Jhevon
Quote:

Originally Posted by Thomas
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:$\displaystyle \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;7 & 3 & 1 & 1 & 5 \;\\\;11 & 9 & 1 & 3 & 17\end{vmatrix}$
Divide R1 by 7: $\displaystyle \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & \frac{3}{7} & 1 & 1 & 5 \;\\\;11 & 9 & 1 & 3 & 17\end{vmatrix}$

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:

$\displaystyle \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;7 & 3 & 1 & 1 & 5 \;\\\;11 & 9 & 1 & 3 & 17 \;\\\; 0 & 0 & 0 & 0 & 0\end{vmatrix}$

now find the reduced row echelon form of that
• Sep 15th 2007, 08:47 PM
Thomas
Quote:

Originally Posted by Thomas
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:$\displaystyle \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;7 & 3 & 1 & 1 & 5 \;\\\;11 & 9 & 1 & 3 & 17\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}$
$\displaystyle \frac{R1}{7} \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & \frac{3}{7} & \frac{1}{7} & \frac{1}{7} & \frac{5}{7} \;\\\;11 & 9 & 1 & 3 & 17\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}$
$\displaystyle R2 - 11\times R1 \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & \frac{3}{7} & \frac{1}{7} & \frac{1}{7} & \frac{5}{7} \;\\\;0 & \frac{30}{7} & \frac{-4}{7} & \frac{10}{7} & \frac{64}{7}\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}$
$\displaystyle R2\times\frac{7}{30} \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & \frac{3}{7} & \frac{1}{7} & \frac{1}{7} & \frac{5}{7} \;\\\;0 & 1 & \frac{-2}{15} & \frac{1}{3} & \frac{32}{15}\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}$
$\displaystyle R1 - \frac{3}{7}R2 \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & 0 & \frac{1}{5} & 0 & \frac{-1}{5} \;\\\;0 & 1 & \frac{-2}{15} & \frac{1}{3} & \frac{32}{15}\;\\\;0 & 0 & 0 & 0 & 0\end{vmatrix}$

Therefore:
$\displaystyle a + \frac{1}{5}c = \frac{-1}{5}$
$\displaystyle b - \frac{2}{15}c + \frac{1}{3}d = \frac{32}{15}$

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.

I finally finished writing my solution. I also added the zero's. Is there anywhere that will write it out for me, instead of me having to write out all the code? It took forever!

Even though I had no zero's when doing it, it still works out the same, right?
• Sep 15th 2007, 08:50 PM
ticbol
Quote:

Originally Posted by Thomas
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:$\displaystyle \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;7 & 3 & 1 & 1 & 5 \;\\\;11 & 9 & 1 & 3 & 17\end{vmatrix}$
$\displaystyle \frac{R1}{7} \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & \frac{3}{7} & \frac{1}{7} & \frac{1}{7} & \frac{5}{7} \;\\\;11 & 9 & 1 & 3 & 17\end{vmatrix}$
$\displaystyle R2\times\frac{7}{30} \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & \frac{3}{7} & \frac{1}{7} & \frac{1}{7} & \frac{5}{7} \;\\\;11 & 9 & 1 & 3 & 17\end{vmatrix}$
$\displaystyle R1 - \frac{3}{7}R2 \begin{array}{c}\ \\ \\ \end{array}\;\begin{vmatrix}\;1 & \frac{3}{7} & \frac{1}{7} & \frac{1}{7} & \frac{5}{7} \;\\\;0 & 1 & \frac{-2}{15} & \frac{1}{3} & \frac{32}{15}\end{vmatrix}$

Therefore:
$\displaystyle a + \frac{1}{5}c = \frac{-1}{5}$
$\displaystyle b - \frac{2}{15}c + \frac{1}{3}d = \frac{32}{15}$

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.

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.
• Sep 15th 2007, 09:01 PM
Thomas
Thanks for the help, ticbol. I, unfortunately, am learning matrices right now. :cool:
• Sep 15th 2007, 09:13 PM
the_philosipher
Thats 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.
• Sep 15th 2007, 10:59 PM
CaptainBlack
Quote:

Originally Posted by the_philosipher
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.

Since you both seem to be in the same city, the chances are it's the same
professor.

RonL
• Sep 15th 2007, 11:01 PM
the_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).
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last