Stuck here

[achi_kun]

 

[Deleted member 4993]

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Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING

Please share your work/thoughts about this problem.

 

[pka]

The question is $\large \dfrac{1}{x}>2$ Please do not over complicate this.
Note that $\large x\ne 0$ so there are two cases.
1) If $x<0$ multiply by $x$ to get $1<2x$ or $\dfrac{1}{2}<x<0$ that is a contradiction. Thus $x\cancel{<\;}0$
2) if $x>0$ multiply by $x$ to get $1>2x$ or $\dfrac{1}{2}>x>0$. Thus we have a solution. SEE HERE

 

[Dr.Peterson]

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I see two different problems there, both called #5. For one, they appear to be leading you through a method of solution, but it is rather confusing, particularly as there is at least one typo.

Which problem are you asking about? What have you tried?

 

[Steven G]

If 0<a<b then 1/b< 1/a

1/x> 2 is the same as 2<1/x so 0<2 < 1/x. Then from the rule above x<1/2

Also you can think about it. You should know that 1/(1/2) = 2 and if you make 1/2 even smaller then the reciprocal will be even larger than 2. So x< 1/2