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 :)
