The distributive property says

A·(B + C) = A·B + A·C

But it also says the reverse

A·B + A·C = A·(B + C) 

Write 13 as 13·1 using the identity property of multiplication
and write 13x as 13·x.

13·1 + 13·x = 13·(1 + x)

That is 

 A·B +  A·C =  A·(B + C)
 ↓ ↓    ↓ ↓    ↓  ↓   ↓ 
13·1 + 13·x = 13·(1 + x)


with 13 substituted for A, 1 substituted for B and x substituted for x.

Edwin