Can inductance or resistance be negative or "infinite"? 
Since the potential different or voltage drop is positive (am I allow to say voltage drop or potential difference in the presence of a changing current?), the current flows from right to left since $\Delta V$ is positive. That is, $V_b$ is at a higher potential than $V_a$.
Case (1), Increasing current.
So traversing from b to a (in the direction of the current) I get a 
$-IR - LI' = 9 \implies -2R - 0.5L = 9$ 
I get $-IR$ because the voltage drops from a higher to a lower potential as it crosses the resistor. I get $-LI'$ because since the current is increasing, the inductor opposes this increase by "becoming" a battery with - on the "a" side and + on the "b" side and I should get a voltage drop across that.
Case (2), Decreasing current.
So traversing from b to a (in the direction of the current) I get a 
$-IR + LI' = 5 \implies -2R - 0.5L = 5$ 
I get $-IR$ because the voltage drops from a higher to a lower potential as it crosses the resistor. I get $LI'$ because since the current is decreasing, the inductor opposes this decrease by "becoming" a battery with + on the "a" side and - on the "b" side and I should get +V by gaining potential.
Solving
http://www.wolframalpha.com/input/?i=RowReduce{{-0.5%2C-2%2C9}%2C{-0.5%2C-2%2C5}}
I get some absurd answer. What is wrong with my argument? 
 A: Your formulas are false. In both case you have : RI+LI'=U.
That gives you : 2R+0.5L=9 and 2R-0.5L=5
So R=3.5 Ohm; L=4 Henry
Notice: The arrow in potential indicates the highest potential Va. Current flows from the highest potential to the lowest as would do a stream down a mountain.
A: $V_{AB} = V_A - V_B$ so node A is more positive than node B side when $V_{AB}>0$
According to the Passive Reference Convention, current enters the positive terminal of a resistor so the current is from node A to node B.
We have
$V_{AB} = 2A \cdot R + 0.5A/s \cdot L = 9V$
and
$V_{AB} = 2A \cdot R - 0.5A/s \cdot L = 5V$
which yields
$R = 3.5 \Omega$
$L = 4 H$
[EDIT]:  Given the numerous questions jak has asked in the comments, I've added a drawing to aid in "seeing" how to get the equations above.
Note that regardless of the direction one "travels" around the loop for KVL, you get the same equation:
$V_{ab} = V_R + V_L = IR + L\frac{dI}{dt} = 2A \cdot R + 0.5A/s \cdot L = 9V$
Also note that I take $V_{ab}$ to be the voltage across nodes a & b with the polarity shown as per the double-subscript convention.  And further, that I chose the current direction and passive component voltage polarities according to the passive reference (or sign) convention.

