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 :)
