anonymous
  • anonymous
3|x| / x-1 < 2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Mertsj
  • Mertsj
b is easy. 3x-2<1 and 3x-2>-1 x<1 and x>1/3
anonymous
  • anonymous
dn't get it..can you please attach?? :(

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anonymous
  • anonymous
its the first question on attachment..please have a look..thanks :)
Mertsj
  • Mertsj
|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.
anonymous
  • anonymous
3 |x| / ( x-1) < 2 this is the question... :'(
anonymous
  • anonymous
humm, I have problem with a and c :(
Mertsj
  • Mertsj
I think just by inspection that x <1 is the solution to the first one.
anonymous
  • anonymous
this question is really confusing :@..
Mertsj
  • Mertsj
The third one, let's just take the log of both sides and solve.
Mertsj
  • Mertsj
x>.79
anonymous
  • anonymous
humm I am trying with log..
Mertsj
  • Mertsj
\[(2x-1)\log_{10}2<(3x-2)\log_{10}3 \]
Mertsj
  • Mertsj
\[(2x-1)(.301)<(3x-2)(.477)\]
anonymous
  • anonymous
are we allowed to use the log value...(if i'm not allowed to bring my calculator then how will i solve).. :(
Mertsj
  • Mertsj
The logs you need would have to be provided.
Mertsj
  • Mertsj
Are calculators forbidden?
anonymous
  • anonymous
I'm not sure..bt, my friend told me.. :'(
anonymous
  • anonymous
thanks man for your efforts... :)
Mertsj
  • Mertsj
yw
Mertsj
  • Mertsj
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.
Mertsj
  • Mertsj
\[\frac{3|x|}{x-1}-2=0\]
Mertsj
  • Mertsj
I found three critical numbers: -2, 2/5 and 1
Mertsj
  • Mertsj
And the inequality is true in all intervals except x>1 so just as I thought, the solution is x<1
anonymous
  • anonymous
hummm...but, its too complicated....are we allowed to use equal sign for inequality? I'm not sure about this process...
Mertsj
  • Mertsj
It is only as an aid to find the critical numbers. It is not part of the solution.
anonymous
  • anonymous
humm got it now..
Mertsj
  • Mertsj
ok
anonymous
  • anonymous
thanks again.. :)
Mertsj
  • Mertsj
yw

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