The first thing I see is the negative sign by the #x#. I don't mess with negative division in inequalities.

Instead, I'll add #3x# to both sides.

#3-3x<=-3#

#3-3x +(3x)<=-3 +(3x)#

We are aloud to do this because we're adding the same quantity to both sides.

Simplified, we get:

#3<=-3+3x#

It has a nice even coefficient now.

Then, we'll get #3x# alone so we can eventually divide.

We'll do this by adding #3# to both sides.

#3+(3)<=-3+3x + (3)#

Simplified, we get:

#6<=3x#

Now, let's divide safely, with no negative sign.

#6 -:(3)<=3x-:(3)#

Simplified, we get:

#2<=x#

Now, it's a little easier to read when #x# is on the left, and that's probably how your teacher wants the answer. We can flip the equation and keep the meaning the same.

#2<=x# becomes #x>=2#

If you don't believe me, just read the first equation forward and backwards. In both instances, #2# is less than or equal to #x#.

Finally, to finish this problem, we should check our work!

Take a number, any number. I usually start with #0# because, well, it's pretty easy to work with.

Plug in #0# to your original equation.

#3-3x<=-3#

#3-3(0)<=-3#

#3<=-3#

So, is #3# really less than (or equal to) #-3#? Obviously not! So before you start getting confused and accuse me of lying to you, let's plug zero into our answer.

#2<=x#

#2<=0#

Is #2# less than (or equal to) #0#? Nope! Since it didn't work in both cases, that means we're probably right. There are two more instances I would try. The first is a number that correctly solves both equations, (so, greater than #2#). The next is the number that #x# could be equal to, (#2#). I'll let you test those, though! Spoiler alert: they work :)