6v + 3(v – 5) = 4v - (2v + 1)
Given = 6v + 3(v — 5) = 4v - (2v + 1)
step 1
when you open the first parenthsesis in step 1
we have 6v + 3v - 15 = 4v - (2v + 1)
Then in step 2
we open the second parenthesis,
This gives;
6v + 3v - 15 = 4v - 2v - 1
Therefore; step 1 and step 2 justified the distributive property