The voltage is higher because the current source and resistance adds to V1.
I solved the first two steps in my head, with no calculator. It should be fairly self-explanatory, just by looking at the schematics.
Step 1, a current source in parallel with a resistor, is equivalent to a voltage source in series with a resistor.
Step 2, the voltages and resistances of the two sources can be added together to form one voltage source and series resistance, as both sources and resistances are in series.
The final step was a bit more tricky.
Calculate the current through R2 and R3, which is the same, as they're in series.
I = V/R
V = V1 - V2 = 11.6V
R = R1 + R2 = 5k + 3.2k = 8.2k
I = 11.6/8200 = 0.0014146 = 1.4146mA
Calculate the voltage across R3:
V = I*R
V = 0.0014146*5 = 0.007073 = 7.073V
Calculate the voltage at node A, which is simply the voltage across R3 added to V2.
V
A = 5 + 7.073 = 12.073V
What's more, the whole circuit can be simplified to a simple voltage source and a series resistor. The voltage is equal to the voltage at node A and the resistance equal to R2 and R3 in parallel.
R1 = (3.2k*5k)/(3.2k+5k) = 16k/8.2k = 1.951k