Just from skimming over it, they probably back-substituted the equations into each other.
Here's how you'd do it in the s-domain:
First, get the loop equations:
V(s) - R1 * I2(s) - L1 * s * I2(s) = 0
V(s) - R2 (I3)s - (1 / (c1 * s)) * I3(s) = 0
And the relation:
I1(s) = I2(s) + I3(s) = 1/s (transform of 1A * u(t))
--note the u(t); this isn't *technically* true, but it works because you're only modeling it from outside the current source - switch "black box"
--also note that we now have three linearly independant equations in three variables (I2, I3, and V). I1 was eliminated as it is already known.
Get the two loop equations equal to each other:
V(s) = R1 * I2(s) + L1 * s * I2(s) = R2 * I3(s) + (1 / (c1 * s)) * I3(s)
--all I did was pull V(s) to one side in both equations, and set them equal to each other.
--now, use the known value of I1 to solve for I2 in terms of I3:
I1(s) = I2(s) + I3(s)
=> I2(s) = 1 / s - I3(s)
--substitute that into the I2 side:
V(s) = R1 * (1 / s - I3(s)) + L1 * s * (1 / s - I3(s)) = R2 * I3(s) + (1 / (c1 * s)) * I3(s)
--and collect terms to get one side with only things multiplied by I3(s):
R1 / s + L1 = R2 * I3(s) + I3(s) / (C1 * s) + R1 * I3(s) + L1 * s * I3(s)
--just pull I3 out as a factor, and divide the other stuff out of the other side to solve for it explicitly:
(R1 / s + L1) / ((R2 + 1 / (C1 * s) + R1 + L1 * s)) = I3(s)
--now that you have I3(s), you repeat the operation from where you solved for I3 in terms of I2 (this step: I2(s) = 1 / s - I3(s)), but solve for I2(s) instead. At this point, you can also solve either equation for V(s), and use that with either eq. to get I2(s).
Also, it might be worthwhile to point out that doing this in the s-domain takes you through exactly the same operations you'd use to solve the linear diff. eq's by hand--the function of s is producing an "auxiliary equation" that you solve algebraically...
*disclaimer: I did the math really fast, and didn't check it afterward...