anonymous
  • anonymous
HELP, PLEASE. 3(x-2)≤1/3 solve inequality
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
this is an "and" case we can set up a 3 part inequality
anonymous
  • anonymous
I am confused what you did there. Did you divide 1/3 to both sides?
anonymous
  • anonymous
correct divide 3 by both sides

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

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