3(z + 6) + 6(z + 9)


In 3(z + 6) there is one multipler outside the parentheses, namely 3, and two
terms inside, namely z and 6.  Multiply the outer 3 by BOTH the inner terms z
and 6 and get 3 times z and 3 times 6.  Three times z is written 3z, and 3
times 6 is, of course, written as 18. So replace 3(z+6) with 3z+18, Now we have

3z + 18 + 6(z + 9)

In 6(z + 9) there is one multipler outside the parentheses, namely 6, and two
terms inside, namely z and 9.  Multiply the outer 6 by BOTH the inner terms z
and 9 and get 6 times z and 6 times 9.  Six times z is written 6z, and 6 times
9 is, of course, written as 54. So replace 6(z+9) with 6z+54, Now we have

3z + 18 + 6z + 54

Now we put like terms together.  The term 3z and the term 6z are like terms
because they both have z's, so we put them together.  The term 18 and the term
54 are like terms because they both have z's, so we put them together.  That
is, we swap places with the 18 and the 6z, and write

3z + 6z + 18 + 54

Now we add 3z and 6z and get 9z. So now we have

9z + 18 + 54

Now we add 18 and 54 and get 72, and this is our final answer

9z + 72

We can go no further because we cannot add unlike terms and get a single term,
because we don't know what z represents. We can only indicate their addition
until someone tells us what number z represents.

Edwin
AnlytcPhil@aol.com