I realize that I'll have to solve this system of simultaneous equations in order to find the values that I need to answer the problem. The problem is that the scale of the problem defeats my ability to do the algebra, so I need to let my calculator do all of the hard work. I have a TI-89, I assume I'll need to edit a matrix of some sort, but I don't know how, with this calculator.

Ok, I took your advice cookie, I tried to enter the equations in the solver (as I'm rusty on my matrixes), here is what I entered:

(for convenience I assigned my difficult to enter variables to single letters easily accessible on the calculator keypad:

I_1 = a
I_2 = b
I_3 = c
I_5 = d
I_6 = e

As well as substituted the values of r's and E's for their numerical values to save typing. As you can see 'a' becomes rather superfluous, I didn't bother to eliminate it from the solve.
)

solve(a = d+e and c= a+b and e = ((200*d + 40)/80) and d = ((20*b+300)/200) and
b = ((200*d-300)/20) and c=((-200*d-80)/70), [variable I'm solving for goes here])

This entry however did not produce the correct results:

b = -6.834 c = -3.476 d = .8166 e = 2.541

The correct answers are:

b = 8 c = 4 d = 1 e = 3

The answers are similar, they are not wildly different (taking into account that the magnitude's of the answers is all that matter with Kirchhoff's rules) and the seeming exactness of the supplied answers makes me suspicious that they were rounded or some such. So can anyone see a flaw in my solve command? Or am I, in fact, correct and the answers supplied are rounded.

I like to use matrix algebra to solve these types of problems.

To put the equations into your TI89 follow this routine:
[[a,b,c][d,e,f][g,h,i][j,k,l]]
These are the constants that multiply your various currents. you should get a four row by three column matrix from this input.

Then enter:
[[w][x][y][z]]
This will give you a column vector. These are the voltage values you get when you sum all the voltages in your given loop, and then move them so they are on the opposite side of the equality from the currents.

Go to "Math" which is the second function on the number five. Go to matrix (4). Go to augment (7). Then you will enter augment([[a,b,c][d,e,f][g,h,i][j,k,l]], [[x][y][z]]), which will give you the four by four matrix you seek. Then go back to matrix and select rref to reduce your matrix. The values on the last column are of course your currents.

This may or may not be an easier method than solving these equations as a system, but it seems more agreeable to me.

I'm going to assume I didn't make any mistakes in modeling the circuit. I've done two other problems similar to this and I'm confident that I've ironed out my misconceptions about modeling the circuits, if not my difficulty with the algebra involved in solving said systems of equations.

from which you can read the solution, in this case, x = 5, y = 2, z = -8.

What Jacob was trying to get you to do was put your variables' coefficients on one side and into one square matrix, and then put what they equal on the other side of the equations and put those into a column matrix, something like

So you're doing fine, except each element in the last column of your matrix is short a negative sign.

Once you have that, call rref() on the matrix you just made. The second argument in rref() is the tolerance, and you don't need to worry about it. Just don't include it.