Solving the Edges first before solving the corners is one of the simplest approaches to the 5E2C-step but also one that rarely produces good results.
Nonetheless, it is still important to learn a few tricks that might help you should this approach prove useful.
There are 3 cases when attempting to solve the edges first.
2 Edges are correctly permuted
3 Edges are correctly permuted
All Edges are correctly permuted
If you happen to get only one edge correctly permuted, turn the U-layer and you'll find that it isn't really one edge at all.
2 Edges Correctly Permuted
Notice that we have to permute only 5edges?
And if 2 of the edges are already in their correct places..
We are left with 3 more edges to permute!
And it's very simple as we can simply cycle the 3 edges with simple side/top moves.
An exmaple:
Notice how the Yellow-Green and Yellow-Orange are correctly permuted.
Also, notice that Yellow-Red, Blue-Red and Yellow-Blue are incorrectly permuted but form a 3-chain cycle.
Using the basic formula (U' R U R'), we can cycle the incorrect edges and get our 5Edges!
3 Edges Correctly Permuted
Keep in mind that we want to permute 5edges.
Now that we have 3 edges correctly permuted, we have only 2 edges left.
Seems easy, right?
Not exactly, 2 edges can't form a cycle.. Only a swap.
And I know not of a swapping commutator.
So, we'll have to make do with TWO 3-edge cycles.
I'll explain the idea behind the two edge-cycles.
First Edge Cycle
In the previous example, Yellow-Orange; Yellow-Blue and Yellow-Red were correctly permuted.
Yellow-Green and Blue-Red were not.
Notice that the Yellow-Green is not in the U-layer.
What we want to do with the first cycle is to make sure that only two edges are solved.
There are 3 conditions but I will only tell you 2 of them first.
They are:
One of the solved edges must be the edge NOT in the U-layer (The Yellow-Green edge piece, in our case)
The solved edges cannot contain an F2L edge (The Blue-Red in our case)
Following these conditions, we are left with 3options:
Yellow-Green && Yellow-Red
Yellow-Green && Yellow-Orange
Yellow-Green && Yellow-Blue
For the sake of speeding things up and telling you the 3rd condition sooner, I shall show you how we will solve Yellow-Green && Yellow-Blue.
Do you see a problem in the above example?
We have solved for Yellow-Green && Yellow-Blue but have accidentally solved for Yellow-Red as well!
We have solved 3 edges when we only wanted to solve 2 and are back to square-one.
Hence, the 3rd condition: The two edges cannot have correct permutations opposite of each other.
Knowing that, we have two options..
Yellow-Green && Yellow-Red
Yellow-Green && Yellow-Orange
After we solve for either, we will be left with 3edges to solve.. Which can be easily done with a simple cycle!