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.