The rule to solving these is to get x on its own by performing 'operations'. However, whatever we do the left-hand side we must also do to the right hand side. So,

, firstly that +9 is messing up the x so we take away 9 from the left to get rid of it but also take away 9 on the right because we
have to.

, next see that the 'times 3' is messing up the x so we divide the left-hand side by 3 to get rid but then also divide the right hand side because we
have to.

and then you have your solution.
Questions like this are all about figuring how to get the x on it's own but also remembering to do whatever you do on
BOTH sides of the equals. Hope this helps.