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b is easy. 3x-2<1 and 3x-2>-1 x<1 and x>1/3
dn't get it..can you please attach?? :(
its the first question on attachment..please have a look..thanks :)
|3x-2|<1 means 3x-2 is less than 1 unit from 0 which means 3x-2 <1 and 3x-2>-1 Just solve each part.
3 |x| / ( x-1) < 2 this is the question... :'(
humm, I have problem with a and c :(
I think just by inspection that x <1 is the solution to the first one.
this question is really confusing :@..
The third one, let's just take the log of both sides and solve.
humm I am trying with log..
are we allowed to use the log value...(if i'm not allowed to bring my calculator then how will i solve).. :(
The logs you need would have to be provided.
Are calculators forbidden?
I'm not sure..bt, my friend told me.. :'(
thanks man for your efforts... :)
for the first one, I found these steps: 1. Move all terms to one side of the inequality sign so that one side is 0. 2. Replace the inequality sign with an equal sign and solve the equation. These solutions are critical values 3. find all values that result in division by 0. These are critical values 4. plot the critical values on a number line. 5. Test each interval defined by the critical values. If an interval satisfies the inequality then it is part of the solution.
I found three critical numbers: -2, 2/5 and 1
And the inequality is true in all intervals except x>1 so just as I thought, the solution is x<1
hummm...but, its too complicated....are we allowed to use equal sign for inequality? I'm not sure about this process...
It is only as an aid to find the critical numbers. It is not part of the solution.
humm got it now..
thanks again.. :)