How to solve distance problems with unequal rates and time.

I've had difficulty with these mostly because my teacher usually stops half way through the problem for us to figure it out for ourselves, which I can't seem to do.
So, Bob goes X-distance down a river/jet-stream/moving sidewalk in Y-amount of time. However, due to the river/jet-stream/moving sidewalk, it takes them Z-extra minutes to make their way back. What was the distance, time, or rate?

Here's an actual example from my book (Saxon Algebra 2 First Edition):
A motor boat makes a 175-kilometer trip down the Mississippi River to New Orleans in 3.3 hours. But it takes an hour longer for the boat to return upstream to its starting point. How fast is the river's current?
And a second slightly different one-
Because of the jet stream, on a round-trip flight from Washington, D.C. to California, the first part of the trip will take half an hour longer than the return trip. How much faster is the plane flying on the return trip?

If you could walk me through both problems in a detailed fashion, that'd be great. :3

In each case, the downstream/tailwind leg of the trip and the upstream/headwind leg of the trips have different rates. You can represent this as:

b + c -- a boat's speed (b) is increased by the speed of the current (c) going downstream

b - c -- a boat's speed (b) is decreased by the speed of the current (c) going upstream

Use these in distance equals rate times time equations:

175 = (b + c) * 3.3 [downstream distance is rate times time]

175 = (b - c) * 4.3 [upstream distance is the slower rate times the longer time]

[simpler method at end of post]

You can multiply out the equations and use them to eliminate b:

(b + c) * 3.3 = 175

becomes

3.3b + 3.3c = 175

and

(b - c) * 4.3 = 175

becomes

4.3b - 4.3c = 175

Multiply both equations by 10 to clear the decimals:

33b + 33c = 1750

43b - 43c = 1750

Multiply the first equation by 43 and the second by 33 to get the b coefficients the same:

1419b + 1419c = 75250

1419b - 1419c = 57750

Subtract equation 2 from equation 1:

1419b + 1419c = 75250
-(1419b - 1419c = 57750)
--------------------------------
2838c = 17500

Divide both sides by 2838:

c = 17500 / 2838 = 6.17 (rounded) km/hr



[added]
There's a simpler way. Divide both sides of equation 1 by 3.3:

b + c = 175 / 3.3 = 53.03 (rounded)

b - c = 175 / 4.3 = 40.70 (rounded)

Subtract equation 2 from equation 1:

b + c = 53.03
-(b - c = 40.70)
------------------
2c = 12.33

Divide both sides by 2:

c = 12.33 / 2 = 6.165 km/hr