diff --git a/SQL/lessons/sql_2.md b/SQL/lessons/sql_2.md
index 2fa2544..76e62e5 100644
--- a/SQL/lessons/sql_2.md
+++ b/SQL/lessons/sql_2.md
@@ -1,6 +1,6 @@
 # Intermediate SQL
 
-## Lesson Objectives tools1. Alter a table
+## Lesson Objectives tools
 
 1. Alter a table
 1. Limit
@@ -10,51 +10,168 @@
 
 ## Alter a table
 
+After you've created a table, you can always make changes
+
+Add a `test` string column
+
+```sql
+ALTER TABLE users ADD COLUMN test VARCHAR(20);
+```
+
+Drop (delete) the test column
+
+```sql
+ALTER TABLE users DROP COLUMN test;
+```
+
+Rename a column:
+
+```sql
+ALTER TABLE users RENAME name TO first_name
+```
+
+Add an id column that increments with each new row (we'll talk about `PRIMARY KEY` later)
+
+```sql
+ALTER TABLE users ADD COLUMN id serial PRIMARY KEY;
+```
+
+Rename a table
+
+```sql
+ALTER TABLE users RENAME TO people;
+```
+
+Change the data type of a column:
+
 ```sql
-ALTER TABLE users ADD COLUMN test VARCHAR(20); -- add an test string column
-ALTER TABLE users DROP COLUMN test; -- drop the test column
-ALTER TABLE users RENAME name TO first_name -- rename a column
-ALTER TABLE users ADD COLUMN id serial PRIMARY KEY; -- add an id column that increments with each new row
-ALTER TABLE users RENAME TO people; -- rename a table
-ALTER TABLE people ALTER COLUMN first_name TYPE text; -- chang the data type of a column
+ALTER TABLE people ALTER COLUMN first_name TYPE text;
 ```
 
 ## Limit
 
+Sometimes it's advantageous to display less than the entire data set:
+
+- If you want to paginate the data for the end user
+- If you have a really massive amount of data, and the queries take a long time to run
+
+Select all rows from people table, but show only the two rows
+
 ```sql
-SELECT * FROM people LIMIT 1; -- select all rows from people table, but show only the first column
-SELECT * FROM people LIMIT 1 OFFSET 1; -- select all rows from people table, but show only one column.  Skip the first row
+SELECT * FROM people LIMIT 2;
+```
+
+Select all rows from people table, but show only the second pair of rows
+
+```sql
+SELECT * FROM people LIMIT 2 OFFSET 2;
 ```
 
 ## Sorting
 
+With sorting, you can begin to start retrieving interesting information from your data
+
+Select all rows from people table, order by name alphabetically:
+
+```sql
+SELECT * FROM people ORDER BY first_name ASC;
+```
+
+Select all rows from people table, order by name reverse alphabetically
+
+```sql
+SELECT * FROM people ORDER BY first_name DESC;
+```
+
+Select all rows from people table, order by age ascending:
+
+```sql
+SELECT * FROM people ORDER BY age ASC;
+```
+
+Select all rows from people table, order by age descending:
+
 ```sql
-SELECT * FROM people ORDER BY first_name ASC; -- select all rows from people table, order by name alphabetically
-SELECT * FROM people ORDER BY first_name DESC; -- select all rows from people table, order by name reverse alphabetically
-SELECT * FROM people ORDER BY age ASC; -- select all rows from people table, order by age ascending
-SELECT * FROM people ORDER BY age DESC; -- select all rows from people table, order by age descending
+SELECT * FROM people ORDER BY age DESC;
 ```
 
 ## Aggregation
 
+Aggregation is another way to retrieve interesting information about our data.  It will place table rows into groups that have the same values for the column you specify.  Then you can perform data analysis on each of those groups
+
+Divide all rows into groups by `first_name`.  Show the `SUM` of the ages of each group.  Also show what `first_name` each group has
+
+```sql
+SELECT SUM(age), first_name FROM people GROUP BY first_name;
+```
+
+Divide all rows into groups by `first_name`.  Show the `AVG` of the ages of each group.  Also show what `first_name` each group has
+
+```sql
+SELECT AVG(age), first_name FROM people GROUP BY first_name;
+```
+
+Divide all rows into groups by `first_name`.  Show the `MAX` of the ages of each group.  Also show what `first_name` each group has
+
+```sql
+SELECT MIN(age), first_name FROM people GROUP BY first_name;
+```
+
+Divide all rows into groups by `first_name`.  Show the `MIN` of the ages of each group.  Also show what `first_name` each group has
+
+```sql
+SELECT MAX(age), first_name FROM people GROUP BY first_name;
+```
+
+Divide all rows into groups by `first_name`.  Show how many rows have a value in the age column.  Also show what `first_name` each group has
+
+```sql
+SELECT COUNT(age), first_name FROM people GROUP BY first_name;
+```
+
+Divide all rows into groups by `first_name`.  Show the number of rows in each group (how many rows have a value in *any* column).  Also show what `first_name` each group has
+
 ```sql
-SELECT SUM(age), first_name FROM people GROUP BY first_name; -- divide all rows into groups by name.  Show the SUM of the ages of each group.  Also show what name each group has
-SELECT AVG(age), first_name FROM people GROUP BY first_name; -- divide all rows into groups by name.  Show the AVG of the ages of each group.  Also show what name each group has
-SELECT MIN(age), first_name FROM people GROUP BY first_name; -- divide all rows into groups by name.  Show the MAX of the ages of each group.  Also show what name each group has
-SELECT MAX(age), first_name FROM people GROUP BY first_name; -- divide all rows into groups by name.  Show the MIN of the ages of each group.  Also show what name each group has
-SELECT COUNT(age), first_name FROM people GROUP BY first_name; -- divide all rows into groups by name.  Show how many rows have a value in the age column.  Also show what name each group has
-SELECT COUNT(*), first_name FROM people GROUP BY first_name; -- divide all rows into groups by name.  Show the number of rows in each group.  Also show what name each group has
-SELECT array_agg(first_name), age FROM people GROUP BY age; -- divide all rows into groups by age.  List the names in each group and show what age each group has
+SELECT COUNT(*), first_name FROM people GROUP BY first_name;
+```
+
+Divide all rows into groups by `age`.  List the `first_name` values in each group and show what `age` each group has
+
+```sql
+SELECT array_agg(first_name), age FROM people GROUP BY age;
 ```
 
 ## JOINS
 
+You can combine tables horizontally
+
+Append each row of the `companies` table onto the end of each row of the `people` table
+
 ```sql
 SELECT * FROM people CROSS JOIN companies;
-SELECT * FROM people JOIN companies ON people.employer_id = companies.id -- find all people who have an employer_id column set and show which company they work for
-SELECT * FROM people LEFT JOIN companies ON people.employer_id = companies.id -- find all people have an employer_id column set and show which company they work for.  In addition to this set, add on all people who do not have an employer_id column set
-SELECT * FROM people RIGHT JOIN companies ON people.employer_id = companies.id -- find all people have an employer_id column set and show which company they work for.  In addition to this set, add on all companies who do not have any people with employer_id columns set to the company's id column
-SELECT * FROM people FULL OUTER JOIN companies ON people.employer_id = companies.id; -- find all people have an employer_id column set and show which company they work for.  In addition to this set, add on all companies who do not have any people with employer_id columns set to the company's id column and all people who do not have an employer_id column set
+```
+
+Do the same, but display only the rows where `people.employer_id` matches `companies.id`.  Note that when you have multiple tables, you'll need to specify which table which column belongs to.  Otherwise, as with the case of `id`, it could be ambiguous which table the column belongs to.
+
+```sql
+SELECT * FROM people JOIN companies ON people.employer_id = companies.id
+```
+
+Do the same as the previous example, but also display any rows from the `people` that were previously left off
+
+```sql
+SELECT * FROM people LEFT JOIN companies ON people.employer_id = companies.id
+```
+
+This is similar to `LEFT JOIN` but it displays any rows from the `companies` table that were previously left off
+
+```sql
+SELECT * FROM people RIGHT JOIN companies ON people.employer_id = companies.id ```
+
+This is basically a combination of `LEFT JOIN` and `RIGHT JOIN`.  Display missing rows from *both* tables
+
+```sql
+SELECT * FROM people FULL OUTER JOIN companies ON people.employer_id = companies.id;
 ```
 
 ## Combining Statments
@@ -62,19 +179,42 @@ SELECT * FROM people FULL OUTER JOIN companies ON people.employer_id = companies
 You can combine `WHERE`, `LIMIT`, `ORDER BY`, `OFFSET` with your `JOIN` statements:
 
 ```sql
-SELECT * FROM people LEFT JOIN companies ON people.employer_id = companies.id WHERE first_name LIKE 'M%' ORDER BY age ASC LIMIT 2 OFFSET 1;
+SELECT * 
+FROM people 
+LEFT JOIN companies 
+	ON people.employer_id = companies.id
+WHERE first_name LIKE 'M%' 
+ORDER BY age ASC 
+LIMIT 2 
+OFFSET 1;
 ```
 
-You can even add `GROUP BY` and aggregation functions too:
+You can even add `GROUP BY` and use aggregation functions too:
 
 ```sql
-SELECT AVG(age), first_name FROM people LEFT JOIN companies ON people.employer_id = companies.id WHERE first_name LIKE 'M%' GROUP BY first_name ORDER BY avg ASC LIMIT 2 OFFSET 1;
+SELECT AVG(age), first_name 
+FROM people 
+LEFT JOIN companies 
+	ON people.employer_id = companies.id 
+WHERE first_name LIKE 'M%' 
+GROUP BY first_name 
+ORDER BY avg ASC 
+LIMIT 2 
+OFFSET 1;
 ```
 
 The order where you add these in the statement matters.  The following breaks (`GROUP BY` must come before `ORDER BY`, `LIMIT`, `OFFSET`):
 
 ```sql
-SELECT AVG(age), first_name FROM people LEFT JOIN companies ON people.employer_id = companies.id WHERE first_name LIKE 'M%' ORDER BY avg ASC LIMIT 2 OFFSET 1 GROUP BY first_name;
+SELECT AVG(age), first_name 
+FROM people 
+LEFT JOIN companies 
+	ON people.employer_id = companies.id 
+WHERE first_name LIKE 'M%' 
+ORDER BY avg ASC 
+LIMIT 2 
+OFFSET 1 
+GROUP BY first_name;
 ```
 
 If it breaks, just try reordering until it works.  You can also research the order in which they need to come in the statement