

So first, let's look at our First case of

1  ×3  +  1  ×2  =  3  +  2  =  5  
2  ×3  3  ×2  6  6  6 
We’re "complicating" (expanding) our fractions by multiplying both numerator and denominator with a same number to get the equivalent pair for our samedenominators sum, i.e. LCD.
After that "transformation" our case becomes simple. We add the numerators,
see if we’ve crossed over to some whole numbers (in this particular case we didn’t 5/6<1), and that’s it.
There was nothing to reduce so we move on to the next combination.
In the following example we see that we don’t need to expand our second fraction, because 4 is already a LCD for 2 and 4.
1  ×2  +  1  =  2  +  1  =  3  
2  ×2  4  4  4  4 
Boring reminder: We don’t add the denominators, just numerators.
Before we move to examples with numerator greater than 1, let's see one more addition  adding 1/4 and 1/6.
1  ×3  +  1  ×2  =  3  +  2  =  5  
4  ×3  6  ×2  12  12  12 
3  ×3  +  2  ×4  =  9  +  8  =  17  =  1  5  
4  ×3  3  ×4  12  12  12  12 
Even with numerators being greater than 1, our addition with unlike denominators hasn’t complicated much. After we find our LCD, in this case 12, we end up with plain addition of numerators.
Here's another example:
5  +  4  ×2  =  5  +  8  =  13  =  2  1  
6  3  ×2  6  6  6  6 
Now some thirds and fourths.
5  ×4  +  5  ×3  =  20  +  15  =  35  =  2  11  
3  ×4  4  ×3  12  12  12  12 
That's the principle. Analogy goes for any other fraction addition. Now all you need is some worksheets to practice (hopefully coming soon).
Anyway, let's summerize below.
Conclusion

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