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this is an "and" case we can set up a 3 part inequality
I am confused what you did there. Did you divide 1/3 to both sides?
correct divide 3 by both sides
what did you get
3(x-2)/3 <=1/3 /3 Would I distribute first and then divide or just divide the whole thing?
You distribute first
I was thinkingg about absolute value sorry
To both sides?
or divide by 3
3x<= 6 1/3
yes what you do to one side you do to the other
Then I would divide by three?
6+1/3 18/3 +1/3 19/3
3x<== 19/3 x<== 19/9
i got x 19/9
Did you turn it into an improper fraction? I am unsure how you got 18/3 + 1/3
yes i turned 6 into an improper fraction
Okay so 19/3 <= X?
x has to bes less than or equal to 19/9
you forgot to divide it by 3
Right, I was about to mention.
So we have 3x => 6 1/3 divided by 3?
so we get x==> 2 1/9
2 1/9 = 19/9
Right, so X<= 19/9 or 2 1/9
no x==> 19/9 we did not divide by a negative to change the sign
So X is greater than or equal to 19/9?
\[x \le 19/9\]
or equal to
Okay, thanks a lot for walking me through it.
Could you help me verify my solution to using some of the numbers from the solution set?
so it has to be less than 2 1/9 lets try 0 since its the easiest
f(0) = 3(0-2)≤1/3 3(-2) ≤1/3 -6≤1/3 its checks out